unit 4 statistical methods btec nationals level 3
DESCRIPTION
BTEC NATIONALS LEVEL 3, MATHS UNIT 4, STATISTICAL METHODSTRANSCRIPT
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Mathematics for Engineering Technicians 29
Statistical Methods
Your view of statistics has probably been formed from what you read in
the papers or what you see on the television Survey use to show which
political party is going to win the election why men grow moustaches if
smoking damages your health the average cost of housing by area and all
sorts of other interesting data So statistics is used to analyse the results of
such surveys and when used correctly it attempts to eliminate the bias that
often appears when collecting data on controversial issues
Statistics is concerned with collecting sorting and analysing numerical
facts which originate from several observations These facts are collated
and summarized then presented as tables charts or diagrams etc
In this brief introduction to statistics we look at two specific areas First we
consider the collection and presentation of data in its various forms Then
we look at how we measure such data concentrating on finding average
values
If you study statistics beyond this course you will be introduced to
the methods used to make predictions based on numerical data and the
probability that your predictions are correct At this stage in your learning
however we will only be considering the areas of data handling and
measurement of central tendency (averages) mentioned above
TYK 410
1 A parallelogram has an area of 60 cm2 if its perpendicular height is 10 cm
what is the length of one of the parallel sides
2Figure 443 shows the cross-section of a template what is its area
3 An annulus has an inside diameter of 075 m and an external diameter of
09 m determine its area
4 Find the volume of a circular cone of height 6 cm and base radius 5 cm
5 Find the area of the curved surface of a cone (not including base) whose
base radius is 3 cm and whose vertical height is 4 cm Hint you need first
to find the slant height
6 If the area of a circle is 7854 mm2 find its diameter to 2 significant figures
7 A cylinder of base radius 5 cm has a volume of 1 L (1000 cm3 ) find its
height
8 A pipe of thickness 5 mm has an external diameter of 120 mm find thevolume of 24 m of pipe material
9 A batch of 2000 ball bearings are each to have a diameter of 5 mm
Determine the volume of metal needed for the manufacture of the whole
batch
10 Determine the volume and total surface area of a spherical shell having an
internal diameter of 6 cm and external diameter of 8 cm
50 cm
30 cm
6 0 c m
3 cm
4 cm
Figure 443 Figure for question 2 in TYK
410
T e s
t yo u r k
n
o w l e
d g e
TYK
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Mathematics for Engineering Technicians300
U N I T 4
Data manipulation
In almost all scientific engineering and business journals newspapers
and Government reports statistical information is presented in the form
of charts tables and diagrams as mentioned above We now look at a
small selection of these presentation methods including the necessary
manipulation of the data to produce them
Charts
Suppose as the result of a survey we are presented with the following
statistical data (Table 44 )
N u m b e r e m p l o y
e d
1000
800
600
400
200
0
P r i v a t e b u s i n e s s
P u b l i c b u s i n e s s
A g r i c u l t u r e
E n g i n e e r i n g
T r a n s p o r t
M a n u f a c t u r e
L e i s u r e i n d u s t r y
E d u c a t i o n
H e a l t h
O t h e r s
Category of employment
Figure 444 Bar chart representing number employed by category
KEY POINT
Statistics is concerned with collecting
sorting and analysing numerical facts
Table 44 Results of a survey
Major category of employment Number employed
Private business 750
Public business 900
Agriculture 200
Engineering 300
Transport 425
Manufacture 325
Leisure Industry 700
Education 775
Health 500
Other 125
Now ignoring for the moment the accuracy of this data let us look at
typical ways of presenting this information in the form of charts in
particular the bar chart and the pie chart
Bar chart
In its simplest form the bar chart may be used to represent data by drawing
individual bars (Figure 444 ) using the figures from the raw data (the data in
the table)
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Mathematics for Engineering Technicians 30
Now the scale for the vertical axis the number employed is easily decided
by considering the highest and lowest values in the table 900 and 125
respectively Therefore we use a scale from 0 to 1000 employees Along
the horizontal axis we represent each category by a bar of even width We
could just as easily have chosen to represent the data using column widths
instead of column heights
Now the simple bar chart above tells us very little that we could not have
determined from the table So another type of bar chart that enables us to
make comparisons the proportionate bar chart may be used
In this type of chart we use one bar with the same width throughout its
height with horizontal sections marked-off in proportion to the whole In
our example each section would represent the number of people employed
in each category compared with the total number of people surveyed
In order to draw a proportionate bar chart for our employment survey we
first need to total the number of people who took part in the survey This
total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If
we were to choose height then we need to set our vertical scale at some
convenient height say 10 cm Then we would need to carry out 10 simple
calculations to determine the height of each individual column
For example given that the height of the total 10 cm represents 5000
people then the height of the column for those employed in private
business 750
500010 1 5
cm This type of calculation is then repeated
for each category of employment The resulting bar chart is shown in
Figure 445
10 cm Others
Health
Education
Leisure industry
Manufacture
Transport
Engineering
Agriculture
Public business
Private business
Figure 445 Proportionate bar chart graduated by height
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Mathematics for Engineering Technicians302
U N I T 4
Example 449
Draw a proportionate bar chart for the employment survey shown in Table 44
using the percentage method
For this method all that is required is to 1047297nd the appropriate percentage of the total
(5000) for each category of employment Then choosing a suitable height of column to
represent 100 mark on the appropriate percentage for each of the 10 employment
categories To save space only the 1047297rst 1047297ve categories of employment have been
calculated
1 private business 750
5000100 15
2 public business 900
5000100 18
3
agriculture 200
5000100 4
4 engineering 300
5000100 6
5 transport 425
5000100 8 5
Similarly manufacture 65 leisure industry 14 education 155 health 10
and other categories 25
Figure 446 shows the completed bar chart
Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One
last type may be used to depict data given in chronological (time) order
Thus for example the horizontal x -axis is used to represent hours days
years etc while the vertical axis shows the variation of the data with time
Example 450
Represent the following data on a chronological bar chart
Year Number employed in general
engineering (thousands)
2003 800
2004 785
2005 690
2006 670
2007 590
Since we have not been asked to represent the data on any speci1047297c bar chart we will use
the simplest involving only the raw data Then the only concern is the scale we should
use for the vertical axis
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Mathematics for Engineering Technicians 30
To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is
rising or falling with time we could use a very much exaggerated scale (Figure 447 b )
This immediately emphasizes the downward trend since 1995 Note that this data is
1047297ctitious (made-up) and used here merely for emphasis
Pie chart
In this type of chart the data is presented as a proportion of the total using
the angle or area of sectors The method used to draw a pie chart is best
illustrated by example
Others (25)
Health (10)
Education (155)
Leisure industry (14)
Manufacture (65)
Transport (85)
Engineering (6)
Agriculture (4)
Public business (18)
Private business (15)
Figure 446 Proportionate percentage bar chart
N u m b e r e m p l o y e
d i n
e n g i n e e r i n g ( t h o u s
a n d s )
1000
900
800
700
600
500
400
300
200
100
02003 2004 2005 2006 2007
Time (years)
(a)
Figure 447 Chronological bar chart (a) in
correct proportion and (b) with graduated
scale
850
800
750
700
650
600
550
500
02003 2004 2005 2006 2007
Time (years)
(b)
N u m b e r e m p l o
y e d i n
e n g i n e e r i n g ( t h o
u s a n d s )
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians300
U N I T 4
Data manipulation
In almost all scientific engineering and business journals newspapers
and Government reports statistical information is presented in the form
of charts tables and diagrams as mentioned above We now look at a
small selection of these presentation methods including the necessary
manipulation of the data to produce them
Charts
Suppose as the result of a survey we are presented with the following
statistical data (Table 44 )
N u m b e r e m p l o y
e d
1000
800
600
400
200
0
P r i v a t e b u s i n e s s
P u b l i c b u s i n e s s
A g r i c u l t u r e
E n g i n e e r i n g
T r a n s p o r t
M a n u f a c t u r e
L e i s u r e i n d u s t r y
E d u c a t i o n
H e a l t h
O t h e r s
Category of employment
Figure 444 Bar chart representing number employed by category
KEY POINT
Statistics is concerned with collecting
sorting and analysing numerical facts
Table 44 Results of a survey
Major category of employment Number employed
Private business 750
Public business 900
Agriculture 200
Engineering 300
Transport 425
Manufacture 325
Leisure Industry 700
Education 775
Health 500
Other 125
Now ignoring for the moment the accuracy of this data let us look at
typical ways of presenting this information in the form of charts in
particular the bar chart and the pie chart
Bar chart
In its simplest form the bar chart may be used to represent data by drawing
individual bars (Figure 444 ) using the figures from the raw data (the data in
the table)
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
Now the scale for the vertical axis the number employed is easily decided
by considering the highest and lowest values in the table 900 and 125
respectively Therefore we use a scale from 0 to 1000 employees Along
the horizontal axis we represent each category by a bar of even width We
could just as easily have chosen to represent the data using column widths
instead of column heights
Now the simple bar chart above tells us very little that we could not have
determined from the table So another type of bar chart that enables us to
make comparisons the proportionate bar chart may be used
In this type of chart we use one bar with the same width throughout its
height with horizontal sections marked-off in proportion to the whole In
our example each section would represent the number of people employed
in each category compared with the total number of people surveyed
In order to draw a proportionate bar chart for our employment survey we
first need to total the number of people who took part in the survey This
total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If
we were to choose height then we need to set our vertical scale at some
convenient height say 10 cm Then we would need to carry out 10 simple
calculations to determine the height of each individual column
For example given that the height of the total 10 cm represents 5000
people then the height of the column for those employed in private
business 750
500010 1 5
cm This type of calculation is then repeated
for each category of employment The resulting bar chart is shown in
Figure 445
10 cm Others
Health
Education
Leisure industry
Manufacture
Transport
Engineering
Agriculture
Public business
Private business
Figure 445 Proportionate bar chart graduated by height
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians302
U N I T 4
Example 449
Draw a proportionate bar chart for the employment survey shown in Table 44
using the percentage method
For this method all that is required is to 1047297nd the appropriate percentage of the total
(5000) for each category of employment Then choosing a suitable height of column to
represent 100 mark on the appropriate percentage for each of the 10 employment
categories To save space only the 1047297rst 1047297ve categories of employment have been
calculated
1 private business 750
5000100 15
2 public business 900
5000100 18
3
agriculture 200
5000100 4
4 engineering 300
5000100 6
5 transport 425
5000100 8 5
Similarly manufacture 65 leisure industry 14 education 155 health 10
and other categories 25
Figure 446 shows the completed bar chart
Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One
last type may be used to depict data given in chronological (time) order
Thus for example the horizontal x -axis is used to represent hours days
years etc while the vertical axis shows the variation of the data with time
Example 450
Represent the following data on a chronological bar chart
Year Number employed in general
engineering (thousands)
2003 800
2004 785
2005 690
2006 670
2007 590
Since we have not been asked to represent the data on any speci1047297c bar chart we will use
the simplest involving only the raw data Then the only concern is the scale we should
use for the vertical axis
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is
rising or falling with time we could use a very much exaggerated scale (Figure 447 b )
This immediately emphasizes the downward trend since 1995 Note that this data is
1047297ctitious (made-up) and used here merely for emphasis
Pie chart
In this type of chart the data is presented as a proportion of the total using
the angle or area of sectors The method used to draw a pie chart is best
illustrated by example
Others (25)
Health (10)
Education (155)
Leisure industry (14)
Manufacture (65)
Transport (85)
Engineering (6)
Agriculture (4)
Public business (18)
Private business (15)
Figure 446 Proportionate percentage bar chart
N u m b e r e m p l o y e
d i n
e n g i n e e r i n g ( t h o u s
a n d s )
1000
900
800
700
600
500
400
300
200
100
02003 2004 2005 2006 2007
Time (years)
(a)
Figure 447 Chronological bar chart (a) in
correct proportion and (b) with graduated
scale
850
800
750
700
650
600
550
500
02003 2004 2005 2006 2007
Time (years)
(b)
N u m b e r e m p l o
y e d i n
e n g i n e e r i n g ( t h o
u s a n d s )
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
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Mathematics for Engineering Technicians 30
Now the scale for the vertical axis the number employed is easily decided
by considering the highest and lowest values in the table 900 and 125
respectively Therefore we use a scale from 0 to 1000 employees Along
the horizontal axis we represent each category by a bar of even width We
could just as easily have chosen to represent the data using column widths
instead of column heights
Now the simple bar chart above tells us very little that we could not have
determined from the table So another type of bar chart that enables us to
make comparisons the proportionate bar chart may be used
In this type of chart we use one bar with the same width throughout its
height with horizontal sections marked-off in proportion to the whole In
our example each section would represent the number of people employed
in each category compared with the total number of people surveyed
In order to draw a proportionate bar chart for our employment survey we
first need to total the number of people who took part in the survey This
total comes to 5000 Now even with this type of chart we may representthe data either in proportion by height or in proportion by percentage If
we were to choose height then we need to set our vertical scale at some
convenient height say 10 cm Then we would need to carry out 10 simple
calculations to determine the height of each individual column
For example given that the height of the total 10 cm represents 5000
people then the height of the column for those employed in private
business 750
500010 1 5
cm This type of calculation is then repeated
for each category of employment The resulting bar chart is shown in
Figure 445
10 cm Others
Health
Education
Leisure industry
Manufacture
Transport
Engineering
Agriculture
Public business
Private business
Figure 445 Proportionate bar chart graduated by height
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians302
U N I T 4
Example 449
Draw a proportionate bar chart for the employment survey shown in Table 44
using the percentage method
For this method all that is required is to 1047297nd the appropriate percentage of the total
(5000) for each category of employment Then choosing a suitable height of column to
represent 100 mark on the appropriate percentage for each of the 10 employment
categories To save space only the 1047297rst 1047297ve categories of employment have been
calculated
1 private business 750
5000100 15
2 public business 900
5000100 18
3
agriculture 200
5000100 4
4 engineering 300
5000100 6
5 transport 425
5000100 8 5
Similarly manufacture 65 leisure industry 14 education 155 health 10
and other categories 25
Figure 446 shows the completed bar chart
Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One
last type may be used to depict data given in chronological (time) order
Thus for example the horizontal x -axis is used to represent hours days
years etc while the vertical axis shows the variation of the data with time
Example 450
Represent the following data on a chronological bar chart
Year Number employed in general
engineering (thousands)
2003 800
2004 785
2005 690
2006 670
2007 590
Since we have not been asked to represent the data on any speci1047297c bar chart we will use
the simplest involving only the raw data Then the only concern is the scale we should
use for the vertical axis
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 516
Mathematics for Engineering Technicians 30
To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is
rising or falling with time we could use a very much exaggerated scale (Figure 447 b )
This immediately emphasizes the downward trend since 1995 Note that this data is
1047297ctitious (made-up) and used here merely for emphasis
Pie chart
In this type of chart the data is presented as a proportion of the total using
the angle or area of sectors The method used to draw a pie chart is best
illustrated by example
Others (25)
Health (10)
Education (155)
Leisure industry (14)
Manufacture (65)
Transport (85)
Engineering (6)
Agriculture (4)
Public business (18)
Private business (15)
Figure 446 Proportionate percentage bar chart
N u m b e r e m p l o y e
d i n
e n g i n e e r i n g ( t h o u s
a n d s )
1000
900
800
700
600
500
400
300
200
100
02003 2004 2005 2006 2007
Time (years)
(a)
Figure 447 Chronological bar chart (a) in
correct proportion and (b) with graduated
scale
850
800
750
700
650
600
550
500
02003 2004 2005 2006 2007
Time (years)
(b)
N u m b e r e m p l o
y e d i n
e n g i n e e r i n g ( t h o
u s a n d s )
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians302
U N I T 4
Example 449
Draw a proportionate bar chart for the employment survey shown in Table 44
using the percentage method
For this method all that is required is to 1047297nd the appropriate percentage of the total
(5000) for each category of employment Then choosing a suitable height of column to
represent 100 mark on the appropriate percentage for each of the 10 employment
categories To save space only the 1047297rst 1047297ve categories of employment have been
calculated
1 private business 750
5000100 15
2 public business 900
5000100 18
3
agriculture 200
5000100 4
4 engineering 300
5000100 6
5 transport 425
5000100 8 5
Similarly manufacture 65 leisure industry 14 education 155 health 10
and other categories 25
Figure 446 shows the completed bar chart
Other categories of bar chart include horizontal bar charts where forinstance Figure 444 is turned through 90deg in a clockwise direction One
last type may be used to depict data given in chronological (time) order
Thus for example the horizontal x -axis is used to represent hours days
years etc while the vertical axis shows the variation of the data with time
Example 450
Represent the following data on a chronological bar chart
Year Number employed in general
engineering (thousands)
2003 800
2004 785
2005 690
2006 670
2007 590
Since we have not been asked to represent the data on any speci1047297c bar chart we will use
the simplest involving only the raw data Then the only concern is the scale we should
use for the vertical axis
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is
rising or falling with time we could use a very much exaggerated scale (Figure 447 b )
This immediately emphasizes the downward trend since 1995 Note that this data is
1047297ctitious (made-up) and used here merely for emphasis
Pie chart
In this type of chart the data is presented as a proportion of the total using
the angle or area of sectors The method used to draw a pie chart is best
illustrated by example
Others (25)
Health (10)
Education (155)
Leisure industry (14)
Manufacture (65)
Transport (85)
Engineering (6)
Agriculture (4)
Public business (18)
Private business (15)
Figure 446 Proportionate percentage bar chart
N u m b e r e m p l o y e
d i n
e n g i n e e r i n g ( t h o u s
a n d s )
1000
900
800
700
600
500
400
300
200
100
02003 2004 2005 2006 2007
Time (years)
(a)
Figure 447 Chronological bar chart (a) in
correct proportion and (b) with graduated
scale
850
800
750
700
650
600
550
500
02003 2004 2005 2006 2007
Time (years)
(b)
N u m b e r e m p l o
y e d i n
e n g i n e e r i n g ( t h o
u s a n d s )
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 816
Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
To present a true representation the scale should start from zero and extend to say800 (Figure 447 a ) If we wish to emphasize a trend that is the way the variable is
rising or falling with time we could use a very much exaggerated scale (Figure 447 b )
This immediately emphasizes the downward trend since 1995 Note that this data is
1047297ctitious (made-up) and used here merely for emphasis
Pie chart
In this type of chart the data is presented as a proportion of the total using
the angle or area of sectors The method used to draw a pie chart is best
illustrated by example
Others (25)
Health (10)
Education (155)
Leisure industry (14)
Manufacture (65)
Transport (85)
Engineering (6)
Agriculture (4)
Public business (18)
Private business (15)
Figure 446 Proportionate percentage bar chart
N u m b e r e m p l o y e
d i n
e n g i n e e r i n g ( t h o u s
a n d s )
1000
900
800
700
600
500
400
300
200
100
02003 2004 2005 2006 2007
Time (years)
(a)
Figure 447 Chronological bar chart (a) in
correct proportion and (b) with graduated
scale
850
800
750
700
650
600
550
500
02003 2004 2005 2006 2007
Time (years)
(b)
N u m b e r e m p l o
y e d i n
e n g i n e e r i n g ( t h o
u s a n d s )
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116
Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
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Mathematics for Engineering Technicians304
U N I T 4
Example 451
Represent the data given in Example 450 on a pie chart
Remembering that there are 360deg in a circle and that the total number employed in
general engineering (according to our 1047297gures) was 800 785 690 670 590 3535
(thousands) then we manipulate the data as follows
Year Number employed in general
engineering (thousands)
Sector angle (to nearest half
degree)
2003 800
800
3535360 81 5
2004 785
785
3535360 80
2005 690
690
3535360 70 5
2006 670
670
3535360 68
2007 590
590
3535360 60
Total 3535 360deg
The resulting pie chart is shown in Figure 448
Other methods of visual presentation include pictograms and ideographs
These are diagrams in pictorial form used to present information to thosewho have a limited interest in the subject matter or who do not wish
to deal with data presented in numerical form They have little or no
practical use when interpreting engineering or other scientific data and
apart from acknowledging their existence we will not be pursuing them
further
Frequency distributions
One of the most common and most important ways of organizing and
presenting raw data is through use of frequency distributions
Consider the data given in Table 45 which shows the time in hours that it
took 50 individual workers to complete a specific assembly line task
2004
2005
2006
20072003
Figure 448 Resulting pie chart for
Example 451 employment in engineering
by year
Table 45 Data for assembly line task
11 10 06 11 09 11 08 09 12 07
10 15 09 14 10 09 11 10 10 11
08 09 12 07 06 12 09 08 07 10
10 12 10 10 11 14 07 11 09 09
08 11 10 10 13 05 08 13 13 08
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
From the data you should be able to see that the shortest time for completion
of the task was 05 hour the longest time was 15 hours The frequency of
appearance of these values is once On the other hand the number of times
the job took 1 hour appears 11 times or it has a frequency of 11 Trying
to sort out the data in this ad hoc manner is time consuming and may lead
to mistakes To assist with the task we use a tally chart This chart simply
shows how many times the event of completing the task in a specific time
takes place To record the frequency of events we use the number 1 in a
tally chart and when the frequency of the event reaches 5 we score through
the existing four 1rsquos to show a frequency of 5 The following example
illustrates the procedure
Example 452
Use a tally chart to determine the frequency of events for the data given on the
assembly line task in Table 45
Time (hours) Tally Frequency
05 1 1
06 11 2
07 1111 4
08 1111 1 6
09 1111 111 8
10 1111 1111 1 11
11 1111 111 8
12 1111 4
13 111 3
14 11 2
15 1 1
Total 50
We now have a full numerical representation of the frequency of events So for example
8 people completed the assembly task in 11 hours or the time 11 hours has a frequency
of 8 We will be using the above information later on when we consider measures of
central tendency
The times in hours given in the above data are simply numbers When data
appears in a form where it can be individually counted we say that it is
discrete data It goes up or down in countable steps Thus the numbers
12 34 86 9 111 130 are said to be discrete If however data is
obtained by measurement for example the heights of a group of people
then we say that this data is continuous When dealing with continuous
data we tend to quote its limits that is the limit of accuracy with which we
take the measurements So for example a person may be 174 05 cm
in height When dealing numerically with continuous data or a large
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116
Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians306
U N I T 4
amount of discrete data it is often useful to group this data into classes or
categories We can then find out the numbers (frequency) of items within
each group
Table 46 shows the height of 200 adults grouped into 10 classes
KEY POINT
The grouping of frequency distributions
is a means for clearer presentation of
the facts
Table 46 Height of adults
Height (cm) Frequency
150ndash154 4
155ndash159 9
160ndash164 15
165ndash169 21
170ndash174 32
175ndash179 45
180ndash184 41
185ndash189 22
190ndash194 9
195ndash199 2
Total 200
The main advantage of grouping is that it produces a clear overall picture of
the frequency distribution In Table 46 the first class interval is 150ndash154
The end number 150 is known as the lower limit of the class interval and
the number 154 is the upper limit The heights have been measured to the
nearest centimetre That means within 05 cm Therefore in effect the
first class interval includes all heights in the range 1495ndash1545 cm these
numbers are known as the lower and upper class boundaries respectivelyThe class width is always taken as the difference between the lower and
upper class boundaries not the upper and lower limits of the class interval
The histogram and frequency graph
The histogram is a special diagram that is used to represent a frequency
distribution such as that for grouped heights shown above It consists of
a set of rectangles whose areas represent the frequencies of the various
classes Often when producing these diagrams the class width is kept the
same so that the varying frequencies are represented by the height of each
rectangle When drawing histograms for grouped data the midpoints of the
rectangles represent the midpoints of the class intervals So for our datathey will be 152 157 162 167 etc
An adaptation of the histogram known as the frequency polygon may also
be used to represent a frequency distribution
Example 453
Represent the data shown in Table 46 on a histogram and draw in the frequency
polygon for this distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116
Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 30
All that is required to produce the histogram is to plot frequency against the height
intervals where the intervals are drawn as class widths
Then as can been seen from Figure 449 the area of each part of the histogram is the product of frequency class width The frequency polygon is drawn so that it connects
the midpoint of the class widths
Another important method of representation is adding all the frequencies
of a distribution consecutively to produce a graph known as a cumulative
frequency distribution or Ogive
Figure 450 shows the cumulative frequency distribution graph for our data
given in Table 46 while Table 47 shows the consecutive addition of the
frequencies needed to produce the graph in Figure 450
From Figure 450 it is now a simple matter to find for example the median
grouped height or as it is more commonly known the 50th-percentile This
occurs at 50 of the cumulative frequency (as shown in Figure 450 ) this
being in our case 100 giving an equivalent height of approximately 175 cm
Any percentile can be found for example the 75th-percentile where in
our case at a frequency of 150 the height can be seen to be approximately
180 cm
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
0
Class width 5 cm
Height of adults in cm
152 157 162 167 172 177 182 187 192 197
Figure 449 Figure for Example 453 histogram showing frequency distribution
KEY POINTThe frequencies of a distribution may
be added consecutively to produce a
graph known as a cumulative frequency
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116
Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1616
Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians308
U N I T 4
C u m u l a t i v e F r e q u e n c y
20
30
40
50
60
70
80
90
100
110
120
130140
150
160
170
180
190
200
10
0
152 157 162 167 172 177 182
180cm
187 192 197
75th percentile
50th percentile
Figure 450 Cumulative frequency distribution graph for data given in Table 46
Table 47 Cumulative frequency data
Height (cm) Frequency Cumulative frequency
150ndash154 4 4
155ndash159 9 13
160ndash164 15 28
165ndash169 21 49
170ndash174 32 81
175ndash179 45 126
180ndash184 41 167
185ndash189 22 189
190ndash194 9 198
195ndash199 2 200
Total 200 200
TYK 411
1 In a particular university the number of students enrolled by a faculty is
given in the table below
Faculty Number of students
Business and administration 1950
Humanities and social science 2820
Physical and life sciences 1050
Technology 850
Total 6670
T e s
t yo u r
k n o
w l e
d g e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1116
Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
httpslidepdfcomreaderfullunit-4-statistical-methods-btec-nationals-level-3 1216
Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
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Mathematics for Engineering Technicians 30
Statistical measurement
When considering statistical data it is often convenient to have one or two
values that represent the data as a whole Average values are often used
You have already found an average value when looking at the median or
50th-percentile of a cumulative frequency distribution So for example we
might talk about the average height of females in the United Kingdom being
170 cm or that the average shoe size of British males is size 9 In statistics
we may represent these average values using the mean median or mode
of the data we are considering We will spend the rest of this short section
finding these average values for both discrete and grouped data starting
with the arithmetic mean
The arithmetic mean
The arithmetic mean or simply the mean is probably the average with whichyou are already familiar For example to find the arithmetic mean of the
numbers 8 7 9 10 5 6 12 9 6 8 all we need to do is to add them all up
and divide by how many there are or more formally
Arithmetic meanarithmetic total of all the individual val
uues
number of values
n
n
sum
where the Greek symbol the sum of the individual values
x 1 x 2 x 3 x 4 middot middot middot x n and n the number of these values in the
data
Illustrate this data on both a bar chart and pie chart
2 For the group of numbers given below produced a tally chart and
determine their frequency of occurrence
36 41 42 38 39 40 42 41 37 40
42 44 43 41 40 38 39 39 43 39
36 37 42 38 39 42 35 42 38 39
40 41 42 37 38 39 44 45 37 40
3 Given the following frequency distribution
Class interval Frequency ( f )
60ndash64 4
65ndash69 11
70ndash74 18
75ndash79 16
80ndash84 7
85ndash90 4
(a) produce a histogram and on it draw the frequency polygon
(b) produce a cumulative frequency graph and from it determine the value
of the 50th-percentile class
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians310
U N I T 4
So for the mean of our ten numbers we have
mean
n
n
sum 8 7 9 10 5 6 12 9 6 8
10
80
108
Now no matter how long or complex the data we are dealing with provided
that we are only dealing with individual values (discrete data) the abovemethod will always produce the arithmetic mean The mean of all the lsquo x
values rsquo is given the symbol x pronounced lsquo x bar rsquo
Example 454
The height of 11 females was measured as follows 1656 cm 1715 cm 1594 cm
163 cm 1675 cm 1814 cm 1725 cm 1796 cm 1623 cm 1682 cm 1573 cm Find
the mean height of these females
Then for n 11
x 165 6 171 5 159 4 163 167 5 181 4 172 5 179 6 162 3 168 22 157 3
1118483
11168 03
x cm
Mean for grouped data
What if we are required to find the mean for grouped data Look back at
Table 46 showing the height of 200 adults grouped into ten classes In this
case the frequency of the heights needs to be taken into account
We select the class midpoint x as being the average of that class and then
multiply this value by the frequency ( f ) of the class so that a value for thatparticular class is obtained ( fx ) Then by adding up all class values in the
frequency distribution the total value for the distribution is obtained ( fx )
This total is then divided by the sum of the frequencies ( f ) in order to
determine the mean So for grouped data
x f x f x f x f x
f f f f
f
f
n n
n
1 1 2 2 3 3
1 2 3
sdot
sdot
sumsum
( )midpoint
This rather complicated looking procedure is best illustrated by
example
Example 455
Determine the mean value for the heights of the 200 adults using the data in
Table 46
The values for each individual class are best found by producing a table using
the class midpoints and frequencies and remembering that the class midpoint is found by
dividing the sum of the upper and lower class boundaries by 2 So for example the mean
value for the 1047297rst class interval is149 5 154 5
2152
The completed table is shown
below
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
Midpoint ( x ) of height (cm) Frequency (f ) fx
152 4 608
157 9 1413
162 15 2430
167 21 3507
172 32 5504
177 45 7965
182 41 7462
187 22 4114
192 9 1728
197 2 394
Total f 200sum fx 35125sum
I hope you can see how each of the values was obtained Now that we have the required
totals the mean value of the distribution can be found
mean value x fx 35125
200175625 05 cm sum
sumf
Notice that our mean value of heights has the same margin of error as the original
measurements The value of the mean cannot be any more accurate than the measured
data from which it was found
Median
When some values within a set of data vary quite widely the arithmetic
mean gives a rather poor representative average of such data Under
these circumstances another more useful measure of the average is the
median
For example the mean value of the numbers 3 2 6 5 4 93 7 is 20 which
is not representative of any of the numbers given To find the median value
of the same set of numbers we simply place them in rank order that is 2
3 4 5 6 7 93 Then we select the middle (median) value Since there are
seven numbers (items) we choose the fourth item along the number 5 as
our median value
If the number of items in the set of values is even then we add together the
value of the two middle terms and divide by 2
Example 456
Find the mean and median value for the set of numbers 9 7 8 7 12 70 68
6 5 8
The arithmetic mean is found as
mean x
9 7 8 7 12 70 68 6 5 8
10
200
1020
This value is not really representative of any of the numbers in the set
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U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians312
U N I T 4
To 1047297nd the median value we 1047297rst put the numbers in rank order that is
5 6 7 7 8 8 912 68 70
Then from the ten numbers the two middle values The 5th and 6th values along are 8
and 8 So the medianvalue
8 8
2 8
Mode
Yet another measure of central tendency for data containing extreme
values is the mode Now the mode of a set of values containing discrete
data is the value that occurs most often So for the set of values 4 4 4 5
5 5 5 6 6 6 7 7 7 the mode or modal value is 5 as this value occurs
four times Now it is possible for a set of data to have more than one
mode For example the data used in Example 462 above has two modes
7 and 8 both of these numbers occurring twice and both occurring more
than any of the others A set of data may not have a modal value at all For
example the numbers 2 3 4 5 6 7 8 all occur once and there is
no mode
A set of data that has one mode is called unimodal data with two
modes is bimodal and data with more than two modes is known as
multimodal
When considering frequency distributions for grouped data the modal
class is that group which occurs most frequently If we wish to find
the actual modal value of a frequency distribution we need to draw a
histogram
Example 457
Find the modal class and modal value for the frequency distribution of the height
of adults given in Table 46
Referring back to Table 46 it is easy to see that the class of heights which occurs most
frequently is 175 ndash 179 cm which occurs 45 times
Now to 1047297nd the modal value we need to produce a histogram for the data
We did this for Example 453 This histogram is shown again here with the modalshown
From Figure 451 it can be seen that the modal value 17825 05 cm
This value is obtained from the intersection of the two construction lines AB and CD The
line AB is drawn diagonally from the highest value of the preceding class up to the top
right-hand corner of the modal class The line CD is drawn from the top left-hand corner
of the modal group to the lowest value of the next class immediately above the modal
group Then as can be seen the modal value is read-off where the projection line meets
the x -axis
KEY POINT
The mean median and mode are
statistical averages or measures
of central tendency for a statistical
distribution
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians 31
F r e q u e n c y
50
45
40
35
30
25
20
15
10
5
C
A
B
D
0Height of adults (cm)
152 157 162 167 172 177 182 187 192 197
Modal value 17825 05 cm
Figure 451 Histogram showing frequency distribution and modal value for height of adults
TYK 412
1 Calculate the mean of the numbers 1765 986 1124 1898 959 and
888
2 Determine the mean the median and the mode for the set of numbers 9 8
7 27 16 3 1 9 4 and 116
3 For the set of numbers 8 12 11 9 16 14 12 13 10 9 find the
arithmetic mean the median and the mode
4 Estimates for the length of wood required for a shelf were as follows
Length (cm) 35 36 37 38 39 40 41 42
Frequency 1 3 4 8 6 5 3 2
Calculate the arithmetic mean of the data5 Calculate the arithmetic mean and median for the data shown in the table
Length (mm) 167 168 169 170 171
Frequency 2 7 20 8 3
6 Calculate the arithmetic mean for the data shown in the table
Length of rivet (mm) 98 99 995 100 1005 101 102
Frequency 3 18 36 62 56 20 5
T e s t yo u
r k
n o
w l e
d g
e
TYK
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present
7172019 UNIT 4 Statistical Methods BTEC NATIONALS LEVEL 3
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Mathematics for Engineering Technicians314
U N I T 4
Elementary Calculus Techniques
Introduction
Meeting the calculus for the first time is often a rather daunting business
In order to appreciate the power of this branch of mathematics we
must first attempt to define it So what is the calculus and what is its
function
Imagine driving a car or riding a motorcycle starting from rest over a
measured distance say 1 km If your time for the run was 25 seconds then
we can find your average speed over the measured kilometre from the
fact that speed distancetime Then using consistent units your average
speed would be 1000 m25 s or 40 ms 1 This is fine but suppose you were
testing the vehicle and we needed to know its acceleration after you had
driven 500 m In order to find this we would need to determine how the
vehicle speed was changing at this exact point because the rate at which
your vehicle speed changes is its acceleration To find things such as rate of
change of speed we can use calculus techniques
The calculus is split into two major areas the differential calculus and the
integral calculus
The differential calculus is a branch of mathematics concerned with finding
how things change with respect to variables such as time distance or speed
especially when these changes are continually varying In engineering we
are interested in the study of motion and the way this motion in machinesmechanisms and vehicles varies with time and the way in which pressure
density and temperature change with height or time Also how electrical
quantities vary with time such as electrical charge alternating current
electrical power etc All these areas may be investigated using the
differential calculus
The integral calculus has two primary functions It can be used to find
the length of arcs surface areas or volumes enclosed by a surface Its
second function is that of anti-differentiation For example we can use the
differential calculus to find the rate of change of distance of our motorcycle
7 Tests were carried out on 50 occasions to determine the percentage of
greenhouse gases in the emissions from an internal combustion engine
The results from the tests showing the percentage of greenhouse gases
recorded were as follows
greenhouse gases present 32 33 34 35 36 37
Frequency 2 12 20 8 6 2
(a) Determine the arithmetic mean for the greenhouse gases present
(b) Produce a histogram for the data and from it find an estimate for the
modal value
(c) Produce a cumulative frequency distribution curve and from it determine
the median value of greenhouse gases present