averages lecture presented on the 21 st of september, 2011

Averages

Lecture presented on the 21st of September, 2011

Where we are

Variables that generate raw data occur at random. So that useful information can be obtained about these variables, as shown in the previous lecture, raw data can be organized by using frequency distributions. Furthermore, organized data can be presented by using various graphs, observed during the last lecture

Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc. Chap 3-3

Describing Data Numerically

Arithmetic Mean

Median

Mode

Describing Data Numerically

Variance

Standard Deviation

Coefficient of Variation

Range

Interquartile Range

Central Tendency Variation

Averages

Statistical methods are used to organize and present data, and to summarize data. The most familiar of these methods is finding averages. For example, one may read that the average salary of top-managers is $5 mln and the salary of a Russian professor is less than $1,000

What is the Russian for “average”?

• существительное: среднее число; средняя величина; убыток от аварии судна; распределение убытка от аварии между владельцами; авария; среднее; среднее арифметическое

• глагол: составлять; равняться в среднем; выводить среднее число; составлять; распределять убытки; усреднять

• прилагательное: средний; обычный; нормальный

What is the Russian for “mean”?

• существительное: середина; среднее; средняя величина;

среднее значение; среднее; среднее число; среднее арифметическое; средство, способ

• глагол: намереваться; иметь в виду; подразумевать; подразумеваться; думать; предназначать; предназначаться; значить; предвещать; иметь значение; означать

• прилагательное: средний; серединный; посредственный;

плохой; слабый; скупой; скаредный; захудалый; жалкий;

убогий; низкий; подлый; нечестный; низкого происхождения; придирчивый; недоброжелательный; бедный; скромный; смущающийся; трудный; неподдающийся

“Average” or “Mean”?

• Analyzing various texts we can summarize that the word “average” is used in general cases, in the discussion of medium size, and the word “mean” is used in special cases dealing with specific formulas and applications to find the average


Measures of Central Tendency

Central Tendency

Mean Median Mode

n

xx

n

1ii

Overview

Midpoint of ranked values

Most frequently observed value

Arithmetic average

Arithmetic Mean

Statistical methods are used to organize and present data, and to summarize data. The most familiar of these methods is finding averages. For example, one may read that the average salary of top-managers is $5 mln and the salary of a Russian professor is less than $1,000

Some quotes from Batueva

In the book “American averages” by Mike Feinsilber and William B. Meed, the authors state:

"Average" when you stop to think of it is a funny concept. Although it describes all of us it describes none of us... While none of us wants to be average American, we all want to know

about him and her”


The authors go on to give examples of averages:

The average American man is five feet, nine inches tall; the average woman is five feet, 3.6

inches. The average American is sick in bed seven

days a year missing five days of work. On the average day, 24 million people receive animal bites


By his or her 7th birthday, the average American will have eaten 14 steers, 1050 chickens, 3.5 lambs, and 25.2 hogs.”

In the above examples, the word “average” is ambiguous, since there are several different methods used to obtain an average. Loosely stated, the average means the center of the distribution or the most typical case. Measures of central tendency are also called measures of average and include the mean, mode and median


The mean is also known as arithmetic mean (or arithmetic average) and is the sum of the values divided by the total number of values in the sample or of the size of population.

Fоr discrete (ungrouped) data the simple mean should be used

x

Logical formula

• The calculation of any average should start with determining a logical formula. Before multiplying, dividing, or adding anything, it is necessary to make the initial ratio of the average, otherwise known as a logical formula

The initial ratio of the average IRA

,A

IRAB

The initial ratio of the average IRA

where A - the amount of the studied events in the population or sample: A is the summary absolute value;

B – the size of population or sample: B is the number of units in the population or sample.

IRA gives us the level of the studied events per unit of population or sample

Examples of IRA

• The average salary shows how much one employee earns.

What do we take in the numerator and denominator of the IRR?A - amount of funds paid to all employees = wages & salaries fund;B - number of employees

Average salary• The salary of the individual employee is

the individual value. Wages & salaries fund is the summary value, and the average salary is an average

Examples of IRA

• The average price shows how much in average this good costs.

What do we take in the numerator and denominator of the IRR? A – the turnover gained from the sale of goods;B - the total quantity of goods sold

Examples of IRA

• The average cost shows how much money was spent per unit of production.

What do we take in the numerator and denominator of the IRR? A – the cost of production;B - the total quantity of goods produced = the quantity of output

Examples of IRA• The average age shows the average

number of years lived by the population under investigation; this indicator concerns not of necessarily animate objects - this may be the average age of cars, students, buildings, chickens, equipment.

What do we take in the numerator and denominator of the IRR?

A – the total number of years;B - the number of surveyed units

Examples of IRA

• Average lifespan: Life expectancy for people , service life, or average age of used equipment -shows the average number of years lived by investigated units, no matter living or inanimate objects they are.

What do we take in the numerator and denominator of the IRR?

A - total number of life (service) years;

B - the number of surveyed units

Logical formula

For a specific economic indicator we may form only one true logical formula

Types of averages

Mathematicians proved that most averages we use, can be expressed in general terms, by the formula of average power

(средней степенной)

,k

k

k n

xx

The averages used in statistics relate to the class of power averages. The general formula of average power is as follows:

_where x k – average power of grade k;

k – the exponent, which determines the type, or

form of the average; х – the values of variants;

n – the number of variants

If k =1, we get the simple arithmetic mean

AM:

;n

xx

if k =2, we get the square mean SM:

;2

n

xxq

in case of k =0, we get the geometric mean GM:

n

n

ii

nng xxxxx ;...

121

if k = -1, we have the harmonic mean HM:

x

nxh 1

Inequality concerning AM, GM, and HM Правило мажорантности

A well known inequality concerning arithmetic, geometric, and harmonic

means for any set of positive numbers is

hgq xxxx

It is easy to remember noting that the alphabetical order of the letters A, G, and H is preserved in the inequality.

The higher exponent in the formula of average power, the greater the value of

the average

Simple arithmetic mean SAM

• Simple arithmetic mean SAM is used when we have different variants without grouping.In the numerator, we find the amount of variants, in the denominator - the number of variants

Example 1.Productivity of 5 workers was 58, 50, 46, 44, 42 products per shift. Determine the average productivity of five workers. In this case, the solution is as follows:

50 46 58 42 4448

5

xx products

n

Simple Arithmetic Mean SAM

In this case, the simple arithmetic mean SAM was used and it could be calculated by the following equation:

where n – the number of values in the sample or the size of population

,ixx

n

Example 2

The miles-per-gallon fuel tests for ten automobiles are given below: 22.2, 23.7, 16.8, 19.7, 18.3, 19.7, 16.9, 17.2, 18.5, 21.0

Find the mean (the average miles-per-gallon)

Example 2

This solution was taken from one text-book, but

I consider it wrong. We need to proceed first to understand how this problem can be solved


Simple Arithmetic Mean SAM

The arithmetic mean (mean) is the most common measure of central tendency

For a population of N values:

For a sample of size n:

Sample sizen

xxx

n

xx n21

n

1ii

Observed

values

N

xxx

N

xμ N21

N

1ii

Population size

Population values

Weighted arithmetic mean

• Weighted arithmetic mean WAM is used for grouped data. This is the most common power-average

Calculating the WAM for a frequency distribution

Example 3

For data in an ungrouped frequency distribution, the mean can be found as shown in the next example

Example 3. The scores for 25 students on a 5-point quiz are shown below. Find the mean (the average score)

Example 3

Score,Number of students,frequency

0 1 0*1=0

1 2 1*2=2

2 6 2*6=12

3 12 3*12=36

4 3 4*3=12

5 1 5*1=5

Total 25 67

i ix ffi

ix i ix f

Example 3

To find the average score it is necessary to calculate the total scores tor all students. For this, the multiplication between the score and the frequency To find the average score it is necessary to calculate the total scores tor all students. For this, the multiplication between the score and the frequency of each class should be made. Then, the sum of these multiplications must be found (see the last line in the table)

i ix f

Example 3

Further, the total sum has to be divided by the number of students:

The average score for 25 students on a 5-point quiz is 2.68 points

672.68 .

25x points

Example 4

The number of machines serviced by one worker, х Number of workers, f

х .f

1 10 10 2 37 74 3 43 129 4 34 136 5 16 80

Total: 140 429

Calculation:

4293,06

140

x fx machins

f

Weighted mean WAM

In two last examples the weighted arithmetic mean was used and it could be calculated by the following formula:

It is also used for data in grouped frequency distribution after transformation into ungrouped frequency distribution by dint of the midpoint.

.i i

i

x fx

f

Example 5

The following frequency distribution shows the amount of money (USD) 100 families spend for gasoline per month.

Find the mean (the average amount of money)

Example 5Classes of

families by the amount of money

Number of families,

Midpoint

24.5-29.5 5 135

29.5-34.5 10 32 320

34.5-39.5 16 37 592

39.5-44.5 32 42 1344

44.5-49.5 27 47 1269

49.5-54.5 6 52 312

54.5-59.5 4 57 228

Total 100 4200

fi

ixi ix f

24.5 29.527

2ix

The midpoint

The procedure of finding the mean for grouped frequency distribution assumes that all of the raw data values in each class are equal to the midpoint of the class. In reality, this is not true, since the average of the raw data values in each class will not be exactly equal to the midpoint

Example 5

However, using this procedure will give an acceptable approximation of the mean, since some values fall above the midpoint and some values fall below the midpoint for each class.

Average amount of money spent for gasoline by family is $42 per month

4200$42.

100i i

i

x fx

f

Productivity, m

Number of

workers, f

x x·f x` x`·f S _ x – x

_ (x – x)2 · f

x` 2· f

Up to 200 3 190 570 -3 -9 3 -63,9 12249,63 27

200-220 12 210 2520 -2 -24 15 -43,9 23126,52 48

220-240 50 230 11500 -1 -50 65 -23,9 28560,50 50

240-260 56 250 14000 0 0 121 -3,9 851,76 0

260-280 47 270 12690 1 47 168 16,1 12182,87 47

280-300 23 290 6670 2 46 191 36,1 29973,83 92

300-320 7 310 2170 3 21 198 56,1 22030,47 63

320 and

more

2 330 660 4 8 200 76,1 11582,42 32

Total: 200 50780 39 140558 359

Example 6

50780253,9

200

x fx m

f

Modification of the WAM formula

• If f is a relative frequency (SR, a share in the population is given), the classic formula of weighted arithmetic mean WAM is not applicable, we can use its modification:

,1

n

xiii dxx


where

;

;

ii

i

i

i

frequency in absolute value

the size of population

fd ff

f


In fact, we multiply variants by SR in shares

Conclusion

For discrete data the simple arithmetic mean should be used. For ungrouped frequency distribution and for grouped frequency distribution the weighted arithmetic mean WAM should be used

Properties of the WAM

The arithmetic mean has the following mathematical properties, which can be used in a task solution. These mathematical properties let us simplify the problem

The product of the arithmetic mean and the sum of frequencies is equal to the

total volume of the studied events in the population (look at the IRA formula):

fxfiii

x

1st property of the WAM

2nd property of the WAM

The sum of deviations between the variants and the mean is equal to zero:

( ) 0;i ix x f

i ii i i i i i

i

x fx f x f x f f

f

0i i i ix f x f

2. The sum of deviations between the variants and the mean is equal to zero

This means, that in AM deviations from the average are mutually repaid

Properties of the arithmetic mean

• The properties 3 to 5 are used to simplify the calculation, when you need to calculate the average of unsuitable values


3d property of the arithmetic mean

If all variants multiply or divide by the same any constant B, the mean will increase or reduce by the same number of times:

11

ii i i

i i

xf x f xB B x

f f B B


4th property of the arithmetic mean

If all variants reduce or increase by definite amount A, then the mean will reduce or increase by the same amount:

( )i i i i i i i i

i i i i

x A f x f Af x f A fx A

f f f f


5th property of the arithmetic mean

If all frequencies multiply and divide by the same any constant k, then the mean will not be changed:

1

.1

ii i i

ii

fx x f

k k xf

fk k

Properties of the arithmetic mean

• If during the calculation of the WAM its properties have been used, we do not get a normal answer, that could be a transformed value. To get the normal WAM, you must make the reverse operations in reverse order

Simplified calculation of the arithmetic mean

for a frequency distribution

is based on properties of the AM

h – the interval width;c – one of the variants near the middle of the distribution (lying in the middle);А – a digital, a maximal multiple for all frequencies

,chf

fxx

;x c

where xh

;A

ff

Productivity, m

Number of

workers, f

x x·f x` x`·f S _ x – x

_ (x – x)2 · f

x` 2· f

Up to 200 3 190 570 -3 -9 3 -63,9 12249,63 27

200-220 12 210 2520 -2 -24 15 -43,9 23126,52 48

220-240 50 230 11500 -1 -50 65 -23,9 28560,50 50

240-260 56 250 14000 0 0 121 -3,9 851,76 0

260-280 47 270 12690 1 47 168 16,1 12182,87 47

280-300 23 290 6670 2 46 191 36,1 29973,83 92

300-320 7 310 2170 3 21 198 56,1 22030,47 63

320 and

more

2 330 660 4 8 200 76,1 11582,42 32

Total: 200 50780 39 140558 359

Example 6

h=20; c=250; f=f'; A=1

3920 250 253,9

200

x fx h c m

f

Example 7The following data represent the grouping of workers by size of payment:

Classes Number of workers,

500-600 10

600-700 15

700-800 20

800-900 25

900-1000 15

1000-1100 10

More than 1100 5

Total: 100

Find the average size of payment, using the mathematical properties of the mean

Chap 3-70

Example 7

Chap 3-71

Transform the grouped frequency distribution into ungrouped frequency distribution calculating the midpoint (column 3). After that, using the property #4, each item could be reduced by the same constant A. Digital A can be anyone, however it is recommended to accept it equal to the variant with maximal frequency (column 4)

Example 7

Chap 3-72

On the basis of the property #3, reduced variants should be divided by the constant B, which can be anyone too. But it is recommended to accept В equal to the width of classes (column 5). The next table shows the procedure of solving the problem

Example 7

Classes,USD

Num-ber of wor-kers,

Mid-point

500-600 10 550 -300 -3 2 -6

600-700 15 650 -200 -2 3 -6

700-800 20 750 -100 -1 4 -4

800-900 25 850 0 0 5 0

900-1000 15 950 100 1 3 3

1000-1100 10 1050 200 2 2 4

More than 1100 5 1150 300 3 1 3

Total: 100 20 -6

fi 850

,

850,

max

ix A

A

because

f

' 'i ix f

,

100

ix A

BB

ix 5if

Example 7

We have now new variants, which can be

symbolized Since changing of the

variants involved changing of the mean, it is necessary to get back the actual magnitude of the mean. Using the property #5, all frequencies were divided by k=5, because 5 is the maximal multiple (column 6)

' .ii

x Ax

B

Example 7

New frequencies are symbolized According to the property #5, changing of frequencies does not entail changing of the mean.

Thus the formula of the mean will look as follows, using the mathematical properties:

The average wage is $820 per worker.

' '

'

6100 850 $820.

20i i

i

x fx B A

f

'.if


Arithmetic Mean

The most common measure of central tendency Mean = sum of values divided by the number of values Affected by extreme values (outliers)

(continued)

0 1 2 3 4 5 6 7 8 9 10

Mean = 3

0 1 2 3 4 5 6 7 8 9 10

Mean = 4

35

15

5

54321

4

5

20

5

104321

Arithmetic Mean

The arithmetic mean, as a single number representing a whole data set, has important advantages. First, its concept is familiar to most people and intuitively clear. Second, every data set has a mean. It is a measure that can be calculated, and it is unique because every data set has one and only one mean. Finally, the mean is useful for performing statistical procedure such as comparing the means from several data sets

Harmonic Mean HM

The harmonic mean HM is defined as the number of values divided by the sum of the reciprocals of each value. The equations of the HM are shown below:

simple harmonic mean:

;1h

i

nx

x

Harmonic Mean HM

weighted harmonic mean:

where Wi=xifi.

,1

ih

ii

Wx

Wx

Harmonic Mean HM

• HM is the reciprocal of the arithmetic mean: one can get the arithmetic mean as number 1, divided by the harmonic mean and vice versa.

• There are simple and weighted HMs. The weighted formula of HM is used more often

Harmonic Mean HM

In the case of HM, frequencies are not known, but we know the total sum of the values.

In fact, the arithmetic mean and the harmonic mean are applied in the same cases, but under different data sets. And so, before the choice of the mean equation it is necessary to construct the logical (economic) formula

HM is applied when the volumes of investigated variants are used as

weights. Sometimes the problem arises: what

formula should be used - the harmonic mean HM or arithmetic mean AM?

The answer is as follows: Fits the formula, in which both the

numerator and denominator would have values with an economic meaning

AM or HM ?

• The tip:If the initial information gives an averaged value (variant) and the denominator of the logical formula, the AM is used.If variant and the numerator of the logical formula are given, the HM is implemented

AM or HM ?

• In other words:

• If the numerator of the IRA is unknown, we’ll use AM.If the denominator of the IRA is unknown, the HM should be used

Enterprise Number of

employees, persons (fi)

Average salary, RUR (xi)

1 540 31046 2 275 31210 3 458 31130

Total: 1273 ?

Example 8

31046 540 31210 275 31130 458

540 275 45831112

i i

i

x fx

f

RUR

Enterprise Monthly fund of wages & salaries,

RUR thousands (wi)

Average salary, RUR (xi)

1 16 764,84 31 046 2 8 582,75 31 210 3 14 257,54 31 130

Total: 39 605,13 ?

Example 9

16764840 8582750 1425754016764840 8582750 14257540

31064 31210 3113031112

i

i

i

wx

wx

RUR

Example 10 A carpenter buys $500 worth of nails at $50 per pound and

$500 worth of nails at $10 per pound. Find the mean (average price of a pound of nails).

For task solution we have to define how the average price can be found. Construct the logical formula expressing the relation between price, and worth:

( )Total worth

Average price price per poundThe number of pounds

Example 10 In this case, the total worth is known, but the number of pounds is not known. Therefore, the

number of pounds could be counted by the ratio between total worth and price:

The average price per pound of nails is $16.67.

The logical equation allows us to correctly choose the mean equation not breaking the relation of economic processes

( )Total worth

Average price price per poundTotal worth

Price per pound

$500 $500 $1000

$16.67.$500 $500 60$50 $10

pounds

Example 11

Below is the data about wages:

#of workshop

Base month Reporting month

Salary per person,

thousand rubles, Xj

Number of personnel, people, fj

Salary per person, thousand rubles, Xj

Wages fund, thousand rubles, Wi

1 11 30 11 275

2 14.5 40 15.2 684

3 16 20 17 408

4 18 10 21 336

Find the average salary per person in each month

Example 11 First, construct the logical formula, describing the relation between salary, wages fund,

and the number of personnel:

Wages fundAverage salary salary per person

The number of personnel

Example 11For the base month Wages fund is not known, but the number of personnel is known.

Wages fund is the total sum of values Wi and could be defined by multiplication between salary per person (variants xi) and the number of personnel (frequencies fi). Thus, logical formula will be as follows:

Wages fundAverage salary


*Salary per person The number of personnel


Example 11 Using this logical formula, the average salary for the base month could be calculated:

To calculate the average salary the weighted arithmetic mean was used:

11*30 14.5* 40 16* 20 18*1014.1 .

30 40 20 10basex thousand rubles

i i

i

x fx

f

Example 11 For the reporting month, Wages fund is known, but the number of personnel is not

known. The number of personnel could be expressed by dividing Wages fund by the salary per person. Thus, the logical formula has been transformed in the following way:

Wages fund Wages fundAverage salary

Wages fundThe number of personnelSalary per person

Example 11 Using this transformation of the initial logical formula, the average salary could

be calculated on the basis of the weighted harmonic mean:

275 684 408 336 170315.5

1 275 684 408 336 11011 15.2 17 21

ireporting

ii

Wx thousand rubles

Wx

Once more AM or HM ?

The accuracy of the choice of the mean equations is rule-based:

If the numerator of the logical formula is

unknown, the arithmetic mean should be used. If the denominator of the logical formula is

unknown the harmonic mean should be used

The geometric mean GM

Sometimes when we are dealing with quantities that change over a period of time, we need to know an average rate of change, such as an average growth rate over a period of several years. In such cases, the arithmetic mean AM and the harmonic mean HM are inappropriate, because they give wrong answers. What we need to find is the geometric mean GM.

The geometric mean GM is defined as the nth root of the product of n values


• The formula of the simple geometric mean GM is:

The geometric mean is useful for finding the average of percentages, ratios, indexes or growth rates

1 21

... .n

n ng n i

i

x x x x x

Example 12

The growth rate of the Living Life Insurance Corporation for the part three years was 35%, 24%, and 18%. Find the average growth rate.

• First, it is necessary to transform the growth rate into growth factor, using the following equation:

1100%

Growth rateGrowth factor

Example 12

• Further the geometric mean can be applied:

The average growth rate is 25.47% per year

3 1.35*1.24*1.18 1.2547 ( 25.47%).

1 2 ...ng nx x x x


• The geometric mean could be applied when the data do to have large spread. Let us suppose, you want to calculate the average winning amount between maximal and minimal winnings amounts.

• n of the geometric mean is more justified:

Example 13

• The geometric mean could be applied when the data do to have large spread. Let us suppose, you want to calculate the average winning amount between maximal and minimal winnings amounts.

Example 13. Minimal winning amount $1000 and maximal winning amount $100,000. Find the average winning amount

Example 13

• The raw data have big difference, and so, the application of the arithmetic mean will be incorrect. In this case, the application of the geometric mean is more justified:

The average winning amount is $10,000

$1000*$100,000 $10,000.gx

Weighted geometric mean WGM

For analysis of time series the weighted geometric mean could be used. The formula is shown below:

The application of the weighted geometric mean will be needed while solving problems on time series

31 21 2 3 ... .i n

k k kk kg nx x x x x

The quadratic (square) mean QM

• A useful mean for physical sciences is the quadratic mean, which it found by taking the square root of the sum of the average of the squares of each value. There arc two kinds of the quadratic mean: the simple quadratic mean and the weighted quadratic mean:

the simple quadratic mean;

2i

q

xx

n

The quadratic (square) mean QM

the weighted quadratic

mean.

The quadratic mean is applied when the measures of dispersion are calculated

2i i

qi

x fx

f

Chronological mean TM

• This mean formula is applied to the number of instant indicators, especially in time series:

12

1...

2

11321

n

xxxxx nn

X


• Take half of the first and last values, plus all values that are in the middle of the series, the amount received divide by “the number of moment indicators minus 1”


TM is widely used in time series analysis, in socio-economic statistics to determine the average population and average size of the fund, as well as other indicators, calculated at certain points in time


• If calculating the average for two moment indicators, the formula for chronological mean TM transforms into the formula of simple arithmetic mean AM:

221

12

1

2

121 xx

Xxx

The universal set of rules for calculating the average

I offer universal set of rules for calculating the average, which discipline students and allow them to choose correctly the necessary mean.

I.Write down a logical formula IRA of the average calculated. Remember that the logical formula does not depend on the initial data, so it is unique, and is appropriate only for the calculation of the required average


II. Compare the logical formula with the initial data. There may be eight cases.

1. The numerator A is unknown, the data are not grouped – the formula of the simple arithmetic mean AM is used.

2. The numerator A is unknown, the data are grouped – use the weighted arithmetic mean WAM formula. If you know the relative frequency in the form of shares, the modified formula MAM of the weighted arithmetic mean is the best way for calculations


3. The denominator B is unknown, the data are not grouped – the simple harmonic mean HM is most suitable formula to calculate the average.

4. The denominator B is unknown, the data are grouped and weights Wi are different – the weighted harmonic mean WHM should be chosen; if the weights Wi coincide we recommend to apply Case #3


5. The data are incomplete, insufficient and not grouped – the simple geometric mean GM formula is necessary and sufficient.

6. The data are incomplete, insufficient, and grouped – deploy the formula of weighted geometric mean WHM.

7. Time series with instant levels and equal time intervals – the simple chronological mean TM is recommended to be applied


8. Time series with instant levels and unequal time intervals - the weighted chronological mean WTM is suitable.

III. Write the appropriate formula.

IV. Plug the initial data into the formula and perform the necessary calculations. V. Specify the unit of measure in the result received and formulate the economic meaning of this numerical answer

Structural averages

Using the average power for the analysis of the distribution is not enough.Structural averages are used for initial analysis of the distribution of units in the population

Structural averages

Out of numerous list of structural averages we’ll discuss mode, median, quartile, decile, and percentile

Mode - the value of the variant occurring in the population the largest number of times. In everyday life the word “mode" actually has the opposite meaning as fashion

Mode Mo

Mode is the most common variant of frequency distribution. For a discrete series this is the value, which corresponds to the highest frequency

Mode Mo

Mode Mo

The mode is the value that is repeated most often in the data set. A data set can have more than one mode or no mode at all.

If we analyze a discrete series and there are several variants with the highest frequency (which is quite rare), then the mode is defined as the arithmetic average of all the modal variants

Mode Mo

The mode is the value that is repeated most often in the data set. A data set can have more than one mode or no mode at all.

The value that occurs most often in a data set is called the mode

The mode can be defined only for ungrouped frequency distribution and grouped frequency distribution. The mode for ungrouped frequency distribution could be defined by sight, by definition, using the most frequency

3Mo

The number of machines serviced by one worker, х

Number of workers, f

х .f

1 10 10 2 37 74

3 43 129

4 34 136 5 16 80

Total: 140 429

Example 4

Example 14The number of books read by each of the 28 students in a

literature class is given below:

Example 14

The number of books read by each of 28 students in a literature class is given below:

Number of books Number of students, frequency

0 2

1 6

2 12 The most frequency

3 5

4 3

Total 28

The most frequency is 12, it means the mode is 2 books. Thus, the most students have read only two books

The mode for grouped frequency distribution can be defined using the following equation. It is used for interval frequency distribution with equal interval widths:

where xMо - lower boundary of the modal class;

hМо - width of the modal class;

f Мо - frequency of the modal class;

f Мо-1 - frequency of the pre-modal class;

f Мо+1 - frequency of the after-modal class

1

1 1

,2

o o

o o

o o o

M Mo M M

M M M

f fM x h

f f f

Productivity, m

Number of

workers, f

x x·f x` x`·f S _ x – x

_ (x – x)2 · f

x` 2· f

Up to 200 3 190 570 -3 -9 3 -63,9 12249,63 27

200-220 12 210 2520 -2 -24 15 -43,9 23126,52 48

220-240 50 230 11500 -1 -50 65 -23,9 28560,50 50

240-260 56 250 14000 0 0 121 -3,9 851,76 0

260-280 47 270 12690 1 47 168 16,1 12182,87 47

280-300 23 290 6670 2 46 191 36,1 29973,83 92

300-320 7 310 2170 3 21 198 56,1 22030,47 63

320 and

more

2 330 660 4 8 200 76,1 11582,42 32

Total: 200 50780 39 140558 359

Example 6

56 50240 20 248

2 56 50 47Mo m

Finding the modal interval

First, it is necessary to define the modal class by definition, using the most frequency. Therefore, the most frequency is 25 and it conforms to the class 800-900 which is detected as the modal interval


The size of payment, USD Number of workers,%

500-600 10

600-700 15

700-800 20

800-900 25

900-1000 15

1000-1100 10

More than 1100 5

Total: 100

Find the mode

Chap 3-130

25 20800 100 $833.3.

(25 20) (25 15)Mo

Thus, the majority of workers have the salary in the amount of $833.3

Mode Mo

• If the modal interval of the first or the last, the missing frequency (or pre-modal or after-modal) is taken to be zero

The mode on a graph

• The mode can be defined using the histogram. For this, it is necessary to select the highest bar, and then connect its right-wing angle with right-wing angle of previous bar. Further, connect left-wing angle of the highest bar with left-wing angle of the next bar. From the point of intersection of two segments, drop a perpendicular on the X-line. The point of intersection of perpendicular and abscissa axis is called the mode

The mode on a graph

• To determine the mode of a discrete series the frequency polygon is drawn. The distance from the vertical axis to the highest point is the graphics mode


Median

In an ordered list, the median is the “middle” number (50% above, 50% below)

Not affected by extreme values

0 1 2 3 4 5 6 7 8 9 10

Median = 3

0 1 2 3 4 5 6 7 8 9 10

Median = 3

Median

The median is the measure of central tendency different from any of means. The median is a single value from the data set that measures the central item in the data. This single item is the middlemost or most central item in the set of numbers. Half of the items lie above this point, and the other half lie below it.

The median is the midpoint of the data array

For calculating the median the data must be ascended or descended by order

Median Me

• Me is the central, “middle” value of a population. Me – the value of a variant located in the middle of the ordered list. Me is the variant, which lies in the middle of the frequency distribution and divides it into two equal parts.

• In the discrete list Me is determined by definition, in the interval frequency distribution – by the formula


Finding the Median

The location of the median:

If the number of values is odd, the median is the middle number If the number of values is even, the median is the average of

the two middle numbers

Note that is not the value of the median, only the

position of the median in the ranked data

dataorderedtheinposition2

1npositionMedian

2

1n

Finding the Median

• If a discrete list contains an odd number of values, Me is the only value, to the right and the left of which there is the same number of values:

2

1 nxMe

Example 15

• Find the median for the ages of seven preschool children. The ages are 2, 3, 4, 2, 3, 5 and 5. First it is necessary to ascend the data: 2, 2, 3, 3, 4, 5, 5. According to the equation, the median is the (7+l)/2=4th item in the array and conforms to 3 years. Thus, you can say, half of the children are under 3 and the other half of them are over 3

Median

For ungrouped data the median can be defined in the following way. If the data set contains an odd number of items, the middle item of the array is the median. If there is an even number of items, the median is the average of the two middle items and can be calculated using the equation from the next slide

Finding the Median

• If a discrete list contains an even number of values, there are two values, to the right and to the left of which there is the same number of values. Me is the arithmetic mean of these two values:

22

2

2xx nn

Me

Example 16

• The ages of ten college students are given below. Find the median: 18, 24, 20, 35, 19, 23, 26, 23, 19, 20. The data set contains the even number of items. Set the data in the ascending order: 18, 19, 19, 20, 20, 23, 23, 24, 26, 35. Using the equation, the median is the (10+1)/2=5.5th item in the data set. In other words, the median lies between the 5th and the 6th items. Thus, the median is:

Me= (20 + 23)/2= 21.5 years

Therefore, half of the students are under 21.5 and the other half of the students are over 21.5 years old

Finding the Median

• For ungrouped frequency distribution the median Me can be defined using the cumulative frequencies iS

The golden rule

Для дискретного ряда медианой является та варианта, для которой накопленная частота впервые превышает половину от суммы частот

For the discrete frequency distribution the median is the value, for which the cumulative frequency for the first time is more than the half of the total frequencies

Example 4

The number of machines serviced by one worker, х

Number of workers, f

S

1 10 10 2 37 47 3 43 90 4 34 124 5 16 140

Total: 140 -

3Me

Example 14

The number of books read by each of 28 students in a literature class is given below:

Number of books Number of students, frequency

Cumulative frequencies, Si

0 2 2

1 6 8

2 5 13

3 12 25

4 3 28

Total 28

Example 14

To locate the middle point, divide n by 2, which gives 28/2=14. Then locate the point where 14 values would fall below and 14 values would fall above. The 14th item falls in the fourth class and conforms to 3 books. Me = 3 books.

It means, half of the students have read less than 3 books and the other half have read more than 3 books


For grouped frequency distribution the median can be defined using the following equation:

where xМе - lower boundary of the median class; hМе - width of the median class; fМе - frequency of the median class; SМе-1 - cumulative frequencies of the class immediately preceding the median class; - the sum of all frequencies (the size of population, the size of sampling)

,2 1

Me

Me

MeMe f

Sf

hxMe

f

Finding the Median

The median class can be defined following the definition and the Golden Rule: we use the cumulative frequencies, in the median interval they are for the first time bigger than the ½ of the sum of all frequencies

Productivity, m

Number of

workers, f

x x·f x` x`·f S _ x – x

_ (x – x)2 · f

x` 2· f

Up to 200 3 190 570 -3 -9 3 -63,9 12249,63 27

200-220 12 210 2520 -2 -24 15 -43,9 23126,52 48

220-240 50 230 11500 -1 -50 65 -23,9 28560,50 50

240-260 56 250 14000 0 0 121 -3,9 851,76 0

260-280 47 270 12690 1 47 168 16,1 12182,87 47

280-300 23 290 6670 2 46 191 36,1 29973,83 92

300-320 7 310 2170 3 21 198 56,1 22030,47 63

320 and

more

2 330 660 4 8 200 76,1 11582,42 32

Total: 200 50780 39 140558 359

Example 6

This means that half of workers have labor productivity which is less

than 252.5 m, while the other half has productivity more than 252.5 m

20065

2240 20 252,5 m.56

Me


The size of payment, USD Number of workers,%

Cumulative frequencies, Si

500-600 10 10

600-700 15 25

700-800 20 45

800-900 25 70

900-1000 15 85

1000-1100 10 95

More than 1100 5 100

Total: 100

Find the median

Chap 3-154

Example 7

First, it is necessary to divide the sum of all frequencies by 2 to find the halfway point: 100/2=50. Further, let us find the class that contains the 50th value. This class is called the median class and it contains the median. The median class is $800-900

Thus, half of workers have the size of payment less than $820 and the

other half of workers havethe size of payment more than $820

10045

2800 100 $820.25

Me

The median can be defined using the ogive. For this, it is recommended to select the point on the Y-line conforming to ½ of all frequencies. From this point the parallel to X-line should be drawn. From the point of intersection of parallel and ogive it is necessary to drop a perpendicular on the abscissa axis. The point of intersection of perpendicular and X-line is named the median. The next slide shows this procedure with the help of cumulative frequency graph

Мо & Ме

• In practical calculations of Mo and Me their values may be far removed from each other. To better reflect the nature of distribution statisticians use other structural averages


Mode

A measure of central tendency Value that occurs most often Not affected by extreme values Used for either numerical or categorical data There may may be no mode There may be several modes

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Mode = 9

0 1 2 3 4 5 6

No Mode


Five houses on a hill by the beach

Review Example

$2,000 K

$500 K

$300 K

$100 K

$100 K

House Prices:

$2,000,000 500,000 300,000 100,000 100,000


Review Example:Summary Statistics

Mean: ($3,000,000/5)

= $600,000

Median: middle value of ranked data = $300,000

Mode: most frequent value = $100,000

House Prices:

$2,000,000 500,000 300,000 100,000 100,000

Sum 3,000,000


Mean is generally used, unless extreme values (outliers) exist

Then median is often used, since the median is not sensitive to extreme values. Example: Median home prices may be

reported for a region – less sensitive to outliers

Which measure of location is the “best”?

The end

•Wishing you all Great Success and Good Luck!

averages lecture presented on the 21 st of september, 2011

Documents

average salary

word average

measures of average

average woman

average day

lecture slide

average american man

median slide