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STATIS

ADDITMATHE

MODU

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IONALMATICS

TICS

LE 13

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1

_____________________________________

CHAPTER 7 : STATISTICS_____________________________________

CONTENTS PAGE

7.0 STATISTICS

7.1 Concept Map 2

7.2 Mean, Mode and Median of Grouped Data 3 ─ 5

7.3 Test Your Self 1 6

7.4 Range and Interquartile Range of

Grouped Data 7 ─ 8

7.5 Test Your Self 2 8 ─ 9

7.6 Past Year Question 10

7.7 Assessment 11 ─ 13

Answers 14

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7.1 CONCEPT MAP

STATISTICS

x


Measures of

2

Central Tendency

Class m =

InterquRanQ3 ─

Q1 =

Histogram


Measures of
Dispersion
artileRange
Mean
geQ

Mode &Modal
Median
1

Q3 =

First Quartile
Third Quartile

3

7.2 MEAN, MODE AND MEDIAN OF GROUPED DATA(Using a formula)

Activity 1: The number of vehicles that pass by a toll plaza from 1 p.m to 2 p.m. for 60concecutive days is shown in the table below.

Number of vehicles Number of days5950 4

6960 10

7970 24

8980 16

9990 6

(a) Calculate the mean of the number of cars.(b) State the modal class.(c) Draw a histogram and estimate the mode of the number of cars from a

histogram .(d) Calculate the median of the number of cars using formula.

Solution :(a)

Number ofvehicles

Number ofdays( f )

Midpoint(x)

fx

5950 4 54.5 218

6960 10 ( ) ( )

7970 24 ( ) 1788

8980 16 84.5 1352

9990 6 94.5 567

f = ( ) )(fx

Mean,)(

)(

f

fxx = 76 . 17 vehicles.

(b) Modal class = 7970

(c)Class boundary Number of days

(frequency)49.5 ─ 59.5 4

1024166

This is the class withthe highest frequency



4

(c) The histogram is shown below

49.5 59.5 69.5 79.5 89.5 99.5 Number of vehicles

Mode = 76 vehicles

Frequency

5

10

15

20

25



(d)Number of

vehiclesNumber of days

(f)Cumulativefrequency

5950 4 4

6960 10 (14)

Step 1 : M

T

Step 2 : M


5

7970 (24) 38

8980 16 ( )

9990 6 ( )

edian class is given by = 30

2

60

2

TTTn

herefore, the median class is 7970 cm.

edian = cf

Fn

Lm

2

= (___)

24

142

60

( __ )

= 76.17 vehicles

L = lower boundary of the medianclass = 69.5

n = 60f

F = cumulative frequency before themedian class = 14

fm = frequency of the median class=24

c = size of the median class= upper boundary ─ lower

boundary

m


6

7.3 Test Yourself 1

For each set of data below, calculate the median by using a formula.

1. The table below shows the age of a group of students.

Age (years) 0 ─ 4 5 ─ 9 10 ─ 14 15 ─ 19 20 ─ 24 25 ─ 29Number of students 10 20 25 32 28 25

Age Frequency Cumulative frequency0 ─ 45 ─ 9

10 ─ 1415 ─ 1920 ─ 2425 ─ 29

Median =

2. The table below shows the weight of a group of babies . 2. The table below shows theweight of a group of babies.

Weight(kg)

3.0 ─ 3.4 3.5 ─ 3.9 4.0 ─ 4.4 4.5 ─ 4.9 5.0 ─ 5.4

Frequency 21 14 25 18 16

Weight (kg) Frequency Cumulative frequency3.0 ─ 3.43.5 ─ 3.94.0 ─ 4.44.5 ─ 4.95.0 ─ 5.4

Median =



7.4 RANGE AND INTERQUARTILE RANGE OF GROUPED DATA( Using a formula)

Activity 2 : The table below shows the speed of 100 vehicles that pass by a small townin a certain period of time.

Speed (km hour─1) Number of vehicles40 ─ 44 845 ─49 1850 ─ 54 1655 ─ 59 2660 ─ 64 2265 ─ 69 10

Calculate(a) the range(b) the interquartile range using formula.

Solution :

(a) Range =2

)(

2

6965

= 67 ─ 42

= 25 km hour─1

(b)Speed Number of

vehicles(f)Cumulativefrequency

40 ─ 44 8 (8)

Step 1

Midpoint of the highest class─ Midpoint of the lowest class


45 ─49 (18) 2650 ─ 54 16 ( )55 ─ 59 26 (68)

Q1

7

60 ─ 64 (22) 9065 ─ 69 10 ( )

: The Q3 class is given by = 25

4

100

4

TTTn

Q1 class = . 45 ─49

Q3


8

Step 2 : Q1 = c

Q

Fn

Q fL

1

1

4

= ( ) +

18

)(4

100

(5)

= 49.22 km hour─1

Step 3 : The Q3 class is given by = 75)100(

4

3

4

3 TTTn

Q3 class = . 60 ─ 64

Step 4 : Q3 = c

Q

Fn

Q fL

3

3

4

3

= 59.5 +

)(

68(___)4

3

(5)

= 61.09 km hour─1

Hence, the interquartile range = Q3 ─ Q1

= ( ) ─ ( )= 11.87 km hour─1

7.5 Test Yourself 2

For each set of data below, find (a) the range (b) the interquartile range by calculation.

LQ1 = lower boundary of the class Q1

class = 44.5

n = 100f

F = cumulative frequency before theQ1 class = 8

fQ1 = frequency of the Q1 class = 18c = class size

= upper boundary ─ lower boundary= 49.5 ─ 44.5

= 5

LQ3 = lower boundary of the classQ1 class = 59.5

n = 100f

F = cumulative frequency beforethe Q3 class = 68

fQ3 = frequency of the Q1 class = 22c = class size

= upper boundary ─ lowerboundary = 64.5 ─ 59.5

= 5



9

1. The table below shows the number of chickens sold over a period of 100 days.

Number of chickens 10 ─ 14 15 ─ 19 20 ─24 25 ─ 29 30 ─ 34

Number of days 20 25 30 15 10

(a)

(b)

2. The table below shows the transport expenses of the workers of an electronic firm to theirworking place.

Expenses(RM)

0.10 ─ 1.00 1.10 ─ 2.00 2.10 ─ 3.00 3.10 ─ 4.00 4.10 ─ 5.00

Number ofworkers

5 9 12 8 6

(a)

(b)



7.6 Past Year Question

SPM 2005 : Paper 2 Question 4

The diagram below is a histogram which represents the distribution of the marks obtainedby 40 pupils in a test.

Number of Pupils

(a)

4

6

8

10

14

12


2

10

0.5 10.5 20.5 30.5 40.5 50.5 Marks

Without using an ogive, calculate the median mark.[3 marks]

0


11

7.7 Assesment

1. Table below shows the frequency distribution of the age of a group of people in atown.

Age 16 ─ 20 21 ─ 25 26 ─ 30 31 ─ 35 36 ─ 40 41 ─ 45Frequency 18 p 15 8 4 1

Given that the median of the age of a group of people is 24.25, find the value of p.[4 marks]

2. Use graph paper to answer this question.The following table shows the distance of 40 workers houses to their place of work.

Distance (km) Numbers ofworkers

1 ─ 2 43 ─ 4 85 ─ 6 147 ─ 8 109 ─ 10 4



12

(a) Draw a histogram and estimate the modal distance from your histogram[4 marks]

(c) Without drawing an ogive, find the median distance. [3 marks]

(c) Calculate the mean distance. [3 marks]



13

3. The table below shows the distribution of the ages of 100 teachers in a secondaryschool.

Age(years) Number of teachers< 30 8< 35 22< 40 42< 45 68< 50 88< 55 98< 60 100

(a) Based on the given table, complete the following :

Ages(years) Frequency25 ─ 29

[2 marks]

(b) Without drawing an ogive, calculate the interquartile range of the abovedistribution. [6 marks]



14

Answers:

Test Yourself 11. Median = 16.84 years

2. Median = 4.19 kg

Test Yourself 21. ( a ) range = 20

( b ) Q1 = 15.5Q3 = 24.5

Interquartile Range = 9 chickens

2. ( a ) range = 4( b ) Interquartile Range = RM 1. 94

SPM 2005

Median = 24.07

Assessment

1. p = 20

2. (a) Mode = 5.7 km

(b) 5.64 km

(c) x =

f

fx=

40

224= 5.6 km

2. (a)Age Frequency Cumulative

frequency25 ─ 29 18 830 ─ 34 14 2235 ─ 39 20 4240 ─ 44 26 6845 ─ 49 20 8850 ─ 56 10 9855 ─ 59 2 100

(b) Q1 = 35.25 yearsQ3 = 46.25 yearsInterquartile Range = 11 years



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16

STATIS

ADDITMATHE

MODU

http://mathsmoz


IONALMATICS

TICS

LE 14

ac.blogspot.com

17

CHAPTER 7 : STATISTICS

CONTENT PAGE

7.1. Concept Map 2

7.2 Standard Deviation And Variance Of Ungrouped Data 3

7.3. Activity 1 5

7.4. Standard Deviation And Variance Of Ungrouped Data( In Frequency Table )

8

7.5. Standard Deviation And Variance Of Grouped Data( With Class Interval )

9

7.6. Activity 2 11

7.7. Activity 3 : SPM Focus Practice 13

7.8. Answers 16



18

7.1 CONCEPT MAP

STATISTICS

MEASURES OF CENTRAL TENDENCY MEASURES OF DISPERSION

Mean

x =n

x

GROUPEDDATA

UNGROUPEDDATA

Standard Deviation

σ = 2

2

)(xn

x

Variance

σ² = 2

2

)(xn

x

Standard Deviation

σ = 2

2

)(xf

fx

Variance

σ² = 2

2

)(xf

fx

UNGROUPEDDATA

GROUPEDDATA

Mean

x =

f

fx



19

7.2: STANDARD DEVIATION AND VARIANCE OF UNGROUPED DATA

Example 1: Calculate the mean, standard deviation and variance of the set of data 15,17, 20, 23, 25.

Solution :

Step 1: Find the sum of squares of x

x 15 17 20 23 25 ∑x = 100x² 225 400 625 ∑x² = 2068

Step 2: Substitute the values into the formulae

Mean, x =n

x=

100= 20

Standard deviation, σ = 2

2

)(xn

x

= 2)(

5

= 3.688

Variance, σ² = (standard deviation)² = ( )² = 13.6

Standard deviation,

σ = 2)(xn

where

2x = sum of squares of x

n = number of data

x = mean =n

x

Variance,

σ² = 2

2

)(xx



20

Example 2: The set of six integers p, q,6, 10, 15 and 18 has a mean of 9 and a standarddeviation of 6. Find(a) the value of p + q,(b) the value of p and of q if q>p.

Solution :

(a) Given that, mean, x = 9,

n

x= 9

9

6

156

qp

p + q + 49 = 9 x 6p + q = 54 – ( )p + q = 5

(b) Given that, standard deviation, σ = 6

2

2

)(xn

x

= 6

2

2

)(xn

x

= 6²

2

222222

96

1815106qp

366

68522

qp

366

68522

qp

+ 81

6

68522 qp

p² + q² + 685 = 117 x 6p² + q² = ( ) – 685

(5 – q)² + q² = 1725 – 10q + q² + q² - 17 = 0

2q² - 10q + ( ) = 0÷2, q² - 5q + 4 = 0

(q – 1)(q – ) = 0(q – 1) = 0 or (q – 4) = 0

q = 1 or q = 4When q = 1, p = 5 – 1 = 4 ( not accepted because it does not satisfy the condition q > p)

When q = 4, p = 5 – 4 = 1 (accepted)

From (a), p + q = 5p = 5 - q



21

7.3 ACTIVITY 1

1. Find the mean, variance and standard deviation of the set of data 3, 8, 5, 7, 9, 6.

2. Find the mean, variance and standard deviation of the set of data 2.3, 2.8, 3.0, 3.2,2.5, 3.9, 3.3.



22

3. The sum of 10 numbers is 180. The sum of the squares of these numbers is 3360.Find the variance of these 10 numbers.

4. Given a set of data (k – 2), k, k, (5k – 2), where k ≠ 0, find the value of k for which themean equals the standard deviation.



23

5. The mean and the standard deviation for the numbers 1, 3, 5, 6 and 8 are 4.6 and 2.42respectively. Find in terms of k

(a) The mean for numbers 1 + k, 3 + k, 5 + k, 6 + k and 8 + k.(b) The standard deviation for the numbers k + 4, 3k + 4, 5k + 4, 6k + 4 and 8k + 4.



24

7.4: STANDARD DEVIATION AND VARIANCE OF UNGROUPED DATA (inFrequencyTable)

Example : A group of 90 typist took a typing test. They were given a document to type.The following table shows the times they took to type.

Time(minutes), x Number of typists, f15 1016 2817 2418 1319 1029 5

Calculate(a) the mean,(b) the standard deviation

Solution :

Time,(x) f fx fx²15 10 150 225016 ( ) 448 716817 24 ( ) 693718 13 234 421219 10 190 ( )20 5 100 2000

f = 90 fx 261762fx

(a) Mean, x =

171530

f

fxminutes

Variance,

σ² = 2

2

)(

f

fx

Standard deviation,

σ =

2

2

)(xfx

wheref = frequency

x = mean =

f

fx



25

(b) Standard deviation,

σ = 2

2

)(xf

fx

= 217

90

= 1.36 minutes

7.5: STANDARD DEVIATION AND VARIANCE OF GROUPED DATA (with classinterval)

Example : The following table shows the distribution of heights of a group of 40 students.

Height(cm) Number of students159 ─ 162 1163 – 166 4167 – 170 11171 – 174 12175 – 178 6179 – 182 4183 – 186 2

Calculate(a) the mean,(b) the standard deviationof the distribution.

Standard deviation,

σ =

2)(xfx

wheref = frequencyx = class midpoint

x = mean =

f

fx

Variance,

σ² =

2

2

)(xfx



26

Solution:

Step 1:

Height(cm) Mid-point, x f fx fx²159 ─ 162 160.5 1 160.5 25760.25163 – 166 164.5 4 ( ) 108241.00167 – 170 ( ) 11 1853.5 312314.75171 – 174 172.5 ( ) 2070.0 357075.00175 – 178 176.5 6 1059.0 ( )179 – 182 ( ) 4 722.0 130321.00183 – 186 184.5 2 369.0 68080.50

40f fx 11887062fx

Step 2: Substitute the values into the formulae

(a) Mean, x =

3.1726892

f

fxcm

(b) Standard deviation,

σ = 2

2

)(xf

fx

= 40

1188706

= 5.51 cm



27

7.6 ACTIVITY 2:

1. Find the mean and variance of the set of data below.

Number of tickets 26 28 30 32Number of days 10 12 15 11

The table above shows the number of tickets sold over a period of 48 days.



28

2. Find the mean and standard deviation of the set of data below.

Score 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34Frequency 6 10 14 12 8

The table above shows the scores of a group of students who fail in a test.



29

7.7 ACTIVITY 3 : SPM FOCUS PRACTICE

1. SPM 2005, PAPER 1. QUESTION 23.

The mean of four numbers is m . The sum of squares of the numbers is 100 and thestandard deviation is 3k.

Express m in terms of k. [3 marks]

2. SPM 2003, PAPER 2. QUESTION 5.

A set of examination marks 654321 ,,,,, xxxxxx has a mean of 5 and a standard

deviation of 1.5.(a) Find

(i) the sum of the marks, x ,

(ii) the sum of the squares of the marks, 2x . [3 marks]

(b) Each mark is multiplied by 2 and then 3 is added to it.Find, for the new set of marks,

(i) the mean,(ii) the variance. [4 marks]



30

3. SPM 2004, PAPER 2, QUESTION 4.

A set of data consists of 10 numbers. The sum of the numbers is 150 and the sum ofthe squares of the numbers is 2472.

(a) Find the mean and variance of the 10 numbers. [3 marks]

(b) Another number is added to the set of data and the mean is increased by 1.Find(i) the value of this number,(ii) the standard deviation of the set of the set of 11 numbers. [3 marks]



4. SPM 2005, PAPER 2. QUESTION 4(b)

The diagram below is a histogram which represents the distribution of the marksobtained by 40 pupils in a test.

Number of Pupils

4

6

8

10

14

12


2

31

0.5 10.5 20.5 30.5 40.5 50.5

(b) Calculate the standard deviation of the distribution.

0


32

7.8 ANSWERS

ACTIVITY 1:

1. x = 6.333

σ² = 3.893

σ = 1.973

2. x = 3

σ² = 0.2457

σ = 0.4957

3. σ² = 12 4. k = 2

5. (a)5

235 k

(b) 2.42k

ACTIVITY 2:

1. x = 29.13σ² = 4.193

2. x = 22.6σ = 6.216

ACTIVITY 3:

1. m = 25 – 9k² 2. (a) (i) 30x

(ii) 5.1632 x

(b) (i) x = 13(ii) σ² = 9

3. (a) x = 15

σ = 22.2

(b) (i) k = 26(ii) σ = 5.494

4. (b) σ = 11.74



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