Measures of Central Tendency

Definition:

A value which is used in this way to represent the distribution is called an „average‟.

Since the average lies in the centre of a distribution, they are called „measures of central

tendency‟. They are also known as „measures of location‟.

Types of Averages:

(a) Arithmetic Mean,

(b) Geometric Mean,

(c) Harmonic Mean,

(d) Median, and

(e) Mode.

(a) Arithmetic Mean (AM):

1. AM is defined as the value obtained by dividing the sum of the values by their

number.

2. It is expressed as follows for sample data:

n

x

n

xxxx n...........21

3. For population data:

N

x

4. The above formulae are for ungrouped data and they cannot be applied to grouped

data. For grouped data, the formula for AM is as follows:

f

fxx or

k

kk

fff

xfxfxfx

.......

.......

21

2211

Where x (all the values) falling in a class are assumed to be equal to the class

mark or mid point of that class. So on the sum of the values in kth

class would be

fkxk , and the sum of values in all the k classes would be equal to fx.

The total number of values is the sum of class frequencies, i.e., f .

Example:

Class Boundaries Frequency

9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Calculate Mean.

Solution:

Class Boundaries f x (Mid-Point) fx 9.5-19.5 5 14.5 72.5

19.5-29.5 8 24.5 196

29.5-39.5 13 34.5 448.5

39.5-49.5 19 44.5 845.5

49.5-59.5 23 54.5 1253.5

59.5-69.5 15 64.5 967.5

69.5-79.5 7 74.5 521.5

79.5-89.5 5 84.5 422.5

89.5-99.5 3 94.5 283.5

99.5-109.5 2 104.5 209

Total 100 5220

f

fxx

2.52100

5220x

Alternate Formulae for Computing Mean:

1. The computation of AM using the grouped data formula is easily provided that

the values x and f are not large.

2. If the values are large, considerable time can be saved by taking deviations from

an assumed or guessed mean.

3. If A is an assumed or guessed mean and D denotes the deviations of x from A

(i.e., D = x – A) then x = A + D.

4. The AM can be expressed as follows:

n

DAx ------- for ungrouped data

f

fDAx ------- for grouped data

Example:

Class Boundaries Frequency

9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f X D = x – A fD 9.5-19.5 5 14.5 -40 -200

19.5-29.5 8 24.5 -30 -240

29.5-39.5 13 34.5 -20 -260

39.5-49.5 19 44.5 -10 -190

49.5-59.5 23 54.5 0 0

59.5-69.5 15 64.5 10 150

69.5-79.5 7 74.5 20 140

79.5-89.5 5 84.5 30 150

89.5-99.5 3 94.5 40 120

99.5-109.5 2 104.5 50 100

Total 100 -230

Although any class mark can be taken as assumed mean, we take the class mark 54.5 as

A, because it corresponds to the largest frequency. See the above table. We have A =

54.5, n = 100 and fD = -230.

f

fDAx

2.52100

)230(5.54x

Second Alternative Method for Computing Mean:

1. If all the class intervals are equal size „h‟, the computation of the mean can be

further simplified using a „coding variable‟ „u‟, where:

h

D

h

Axu

h

Axu , huAx

hn

uAh

n

uAx -------------------- for ungrouped data

hf

fuAh

f

fuAx ---------------------- for grouped data

2. This method is also known as „coding method‟.

Example:

Class Boundaries Frequency

9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f X D = x – A h

Du fu

9.5-19.5 5 14.5 -40 -4 -20

19.5-29.5 8 24.5 -30 -3 -24

29.5-39.5 13 34.5 -20 -2 -26

39.5-49.5 19 44.5 -10 -1 -19

49.5-59.5 23 54.5 0 0 0

59.5-69.5 15 64.5 10 1 15

69.5-79.5 7 74.5 20 2 14

79.5-89.5 5 84.5 30 3 15

89.5-99.5 3 94.5 40 4 12

99.5-109.5 2 104.5 50 5 10

Total 100 -23

A = 54.5; n = 100; h = 10 and fu = -23

hf

fuAh

f

fuAx

2.5210100

235.54x

Weighted Arithmetic Mean (WAM):

1. WAM is used to find average of certain values which are not of equal importance.

2. The numerical values are called „weights‟, and denoted as w1 , w2 , ….. wk . 3. WAM is expressed as follows:

k

kk

wwww

xwxwxwx

.......

.........

21

2211

w

wxxw

Example:

Sectors Expenditure

(All figures in Rs. Billion) Weight

General Public Service 503 46

Development 272 25

Defence 223 20

Public Order Safety 19 3

Education 17 3

Health 4 2

Housing 1 1

Environment Protection 1 1

Calculate Weighted Average Mean.

Solution:

Sectors X W WX

General Public Service 503 46 23138

Development 272 24 6528

Defence 223 20 4460

Public Order Safety 19 3 57

Education 17 3 51

Health 4 2 8

Housing 1 1 1

Environment Protection 1 1 1

Total 100 34244

w

wxxw

billionRsxw 44.342.100

34244

Properties of Arithmetic Mean:

1. The sum of deviations of values from their mean is equal to zero:

0)( xx or 0)( xxf

2. If n1 values have mean 1x ; n2 values have mean 2x ; and so on nk values have

mean kx , the mean of all the values is:

k

kk

nnn

xnxnxnx

.......

.........

21

2211

n

nxx

3. The sum of squares of the deviation of the values x from any value a is minimum

if and only if a = x :

2)( ax is a minimum, if a = x .

4. The AM is affected by change of origin and scale. By this, we mean that if we

add or subtract a constant from all values or multiply or divide all the values by a

constant, the mean is affected by these changes:

If x = a (a constant), then x = a

If y = x ± a, then y = x ± a

If y = bx, then y = b x

If y = a

x, then y =

a

x

(b) Geometric Mean (GM):

1. GM is defined only for non-zero positive values. It is the nth root of the product

of n values in the data.

2. It can be expressed as follows:

nn xxGM1

)(

Where x = x1 × x2 × x3 × ………. × xn.

Alternate Method for Computing Geometric Mean:

1. In practice, it is difficult to extract higher roots. The GM is, therefore, computed

using logarithms:

n

xxxGM n )log.........log(log

log 21

n

xGM

loglog

n

xGM

logAntilog

2. The GM is used mainly to find the average of ratios, rates of change, economic

indices and the like. It is preferred when data array follows a pattern of

„geometric progression‟.

Example:

Find the GM of the following data:

5,6,7,8,2,3,1,10,13,11

Solution:

nn xxGM1

)(

Where x = x1 × x2 × x3 × ………. × xn.

20.5

1441440014414400)(

1113101328765)1113101328765(

10101

10101

GM

orGM

orGM

Geometric Mean for Grouped Data: For grouped data, the GM is computed as below:

n f

k

ff kxxxGM ........21

21

nf

k

ff kxxxGM1

21 )........( 21

Taking logarithm of both sides:

f

xfxfxfGM kk log........loglog

log 2211

f

xfGM

loglog

f

xfGM

logAntilog

Weighted Geometric Mean:

w

xwGM

logAntilog

Example:

Class Boundaries f 9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f x log x f log x 9.5-19.5 5 14.5 1.1614 8.07

19.5-29.5 8 24.5 1.3892 11.1136

29.5-39.5 13 34.5 1.5378 19.9914

39.5-49.5 19 44.5 1.6484 31.3196

49.5-59.5 23 54.5 1.7364 39.9372

59.5-69.5 15 64.5 1.8096 27.144

69.5-79.5 7 74.5 1.8722 13.1054

79.5-89.5 5 84.5 1.9269 9.6345

89.5-99.5 3 94.5 1.9754 5.9262

99.5-109.5 2 104.5 2.0191 4.0382

Total 100 170.2801

f

xfGM

logAntilog

44.50 1.702801 Antilog

100

2801.170Antilog

GM

GM

(c) Harmonic Mean (HM): HM is defined only for non-zero positive values. It is the reciprocal of mean of

reciprocals of values. More briefly, HM, of a set of n values x1, x2, ….. , xn, is the

reciprocal of the AM of the reciprocals of the values. Thus:

n

xxxHM n

1........11

ofReciprocal 21

n

xxxHM

n

1........11

1

21

nxxx

nHM

1........1121

x

nHM

1 -------------- for ungrouped data

Harmonic Mean for Grouped Data:

The reciprocal of the class marks (in case of grouped data) will be 1

1

x,

2

1

x,…….,

kx

1.

Since the reciprocals occur with frequencies f1, f2, ….. , fk, the total value of the

reciprocals in the first class is 1

1

x

f, in second class

2

2

x

f, ….. , and in the k

th class is

k

k

x

f.

The sum of reciprocals in all the k classes would be:

k

k

x

f

x

f

x

fHM ......

2

2

1

1

x

f

fHM

xf

fHM

1

Weighted Harmonic Mean:

)(x

w

wHM

Example:

Find HM of the values 1, 2 and 3.

Solution:

x

nHM

1

64.1

31

21

11

3HM

Example:

Class Boundaries Frequency

9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f x x

1

xf

1

9.5-19.5 5 14.5 0.06897 0.34485

19.5-29.5 8 24.5 0.04082 0.32656

29.5-39.5 13 34.5 0.02899 0.37687

39.5-49.5 19 44.5 0.02247 0.42693

49.5-59.5 23 54.5 0.01835 0.42205

59.5-69.5 15 64.5 0.01550 0.2325

69.5-79.5 7 74.5 0.01342 0.09394

79.5-89.5 5 84.5 0.01183 0.05915

89.5-99.5 3 94.5 0.01058 0.03174

99.5-109.5 2 104.5 0.00957 0.01914

Total 100 2.33373

xf

fHM

1

85.4233373.2

100HM

Relation between AM, GM and HM:

1. The AM is greater than GM, the GM is greater than HM:

HMGMAM

2. AM, GM and HM are equal when all the values are equal, (e.g., 5,5,5,5,….):

HMGMAM

3. Therefore, the relationship between AM, GM and HM is expressed as follows:

HMGMAM

(d) Median:

1. Median is defined as the middle value of the data when the values are arranged in

ascending or descending order.

2. If there are even number of values in the data, the average of two middle values in

the array is taken as the median:

median theis value2

1or value

2th

nth

N

Median for Grouped Data:

cff

f

hlx

2

~

Where l = lower class boundary

h = width

f = frequency of the median class

cf = cumulative frequency of the preceding class

Example:

Class Boundaries Frequency

9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f x cf 9.5-19.5 5 14.5 5

19.5-29.5 8 24.5 13

29.5-39.5 13 34.5 26

39.5-49.5 19 44.5 45

49.5-59.5 23 54.5 68

59.5-69.5 15 64.5 83

69.5-79.5 7 74.5 90

79.5-89.5 5 84.5 95

89.5-99.5 3 94.5 98

99.5-109.5 2 104.5 100

Total 100

502

100

2

fth value lies in the 5

th class, viz., 50-59 or 49.5-59.5. Therefore, it is the

median class. Here, l = 49.5, h = 10, f = 23, f = 100, and cf = 45.

cff

f

hlx

2

~

67.51452

100

23

105.49~x

Median for Discrete Data:

To find the x~ from discrete data, we form a cumulative frequency. The x~ is the value

corresponding to CF distribution in which 2

)1(nth value lies:

Example:

No. of children No. of families Cumulative frequency

0 4 4

1 25 29

2 53 82

3 18 100

4 14 114

5 6 120

120

Solution:

Since 2

)1(nth value (i.e., 60

2

120th value lies in the CF corresponding to 2, the

median is 2. ( x~ = 2).

Graphical Location of Median:

The approximate value of the median can be located from an ogive, i.e., a cumulative

frequency polygon:

Example:

Class Boundaries f CF

9.5-19.5 5 5

19.5-29.5 8 13

29.5-39.5 13 26

39.5-49.5 19 45

49.5-59.5 23 68

59.5-69.5 15 83

69.5-79.5 7 90

79.5-89.5 5 95

89.5-99.5 3 98

99.5-109.5 2 100

Total 100

Solution:

Quartiles, Deciles and Percentiles:

1. The values which divide an arrayed set of data into four equal parts are called

„quartiles‟.

2. The first and third quartiles are also known as lower and upper quartiles

respectively.

3. The quartiles are expressed as follows:

0

20

40

60

80

100

9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 109.5

Cu

mu

lati

ve F

req

uen

cie

s

Class Boundaries

th value2

f

Median

itemth 4

)1( of Value 1

nQ

itemth 2

)1(or itemth

4

)1(2 of Value 2

nnQ

*

itemth 4

)1(3 of Value 3

nQ

†

4. The values which divide an arrayed set of data into ten equal parts are called

„deciles‟.

itemth 10

)1( of Value 1

nD

itemth 5

)1(or itemth

10

)1(2 of Value 2

nnD

itemth 10

)1(9 of Value

nDg

5. The values which divide an arrayed set of data into one hundred equal parts are

called „percentiles‟:

itemth 100

)1( of Value 1

nP

itemth 50

)1(or itemth

100

)1(2 of Value 2

nnP

itemth 100

)1(99 of Value

nPg

6. The quartiles, deciles and percentiles may be determined from the grouped data in

the same way as the median except that in place of 2

n, we will use

4

f,

10

f

and 100

f:

* It should be noted here that xQ ~

2 .

† The relationship between 321 and , QQQ is expressed as 321 QQQ .

CFf

f

hlQ

41

Where l = lower class boundary of Q1 class

h = width of class boundary

f = frequency of Q1 class

CF = Cumulative frequency of the class

preceding to Q1 class

CFf

f

hlQ

4

33

CFf

f

hlD

101

CFf

f

hlP

1001

7. For discrete data, the quartiles, deciles and percentiles are determined in the same

way as the median.

8. The quartiles, deciles and percentiles may be located from an ogive in a similar

way as the median:

Example:

(See the former example)

Solution:

(See solution next page)

valuethf

Q 254

100

41

valuethf

Q 754

)100(3

4

33

valuethf

D 4010

400

10

44

valuethf

D 7010

700

10

77

valuethf

P 37100

3700

100

3737

valuethf

P 57100

5700

100

5757

(e) Mode:

1. The mode is defined as that value in the data which occurs the greatest number of

times provided that such a value exists.

0

20

40

60

80

100

9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 109.5

Cu

mu

lati

ve F

req

uen

cie

s

Q1 P37 P57 D7

D4 Q3

2. If each value occurs the same number of times, then there is no mode. If two or

more values occurs the same number of times but more frequently than any of the

other values, then there is more than one mode.

3. The distribution having only one mode is called „uni-modal distribution‟, two

modes „bi-modal distribution‟, and more than two modes „multi-modal

distribution‟.

Mode for Grouped Data:

In case of grouped data, the mode is defined as that value of x which corresponds to the

highest points on the curve. The mode is denoted by x̂ (read as “x caret”):

21

1

2ˆ

fff

ffhlx

m

m

Where fm is the frequency of modal class, f1 is the frequency of preceding class, and f2 is

the frequency of following class.

Example:

Class Boundaries f 9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

Solution:

Class Boundaries f 9.5-19.5 5

19.5-29.5 8

29.5-39.5 13

39.5-49.5 19

49.5-59.5 23

59.5-69.5 15

69.5-79.5 7

79.5-89.5 5

89.5-99.5 3

99.5-109.5 2

Total 100

In the above table, the frequency of 5th

class is maximum which is, therefore, the modal

class. Here l = 49.5; h = 10; fm = 23; f1 = 19; and f2 = 15.

21

1

2ˆ

fff

ffhlx

m

m

83.521519)23(2

1923105.49x̂

Mode for Discrete Data:

In case of discrete data, the mode may be picked out by inspection. It is the most

common value, i.e., the value with greatest frequency.

Relation between Mean, Median and Mode:

1. In Symmetrical distribution, the mean, median and mode coincide, i.e., equal in

value.

2. In moderately skewed distributions:

)~(3ˆ xxxx or xxx 2~3ˆ

↑ xxx ˆ~ X

Y

O

x̂ x~ x X

Y

O x x~ x̂ X

Y

O

Advantages and Disadvantages of Averages:

Arithmetic Mean:

Advantages:

1. The arithmetic mean is rigidly defined by a mathematical formula.

2. It is most widely used and most commonly understood of all the averages.

3. It is easy to calculate and is determinate in almost every case.

4. It depends on all the values of the data and a change in any value changes the

value of the mean.

5. It is capable of further algebraic manipulation.

6. It is relatively stable measure.

Disadvantages:

1. The mean is greatly influenced by extreme values especially by extremely large

ones.

2. It is not an appropriate average for highly skewed distributions, e.g., distributions

of wages or incomes, etc., and U-shaped distributions.

3. It cannot be accurately computed in case of an open-end frequency table.

4. It may locate the value at a point at which few or none or the actual observations

lie.

Geometric Mean:

Advantages:

1. The GM is rigidly defined by a mathematical formula.

2. It is based on all the values.

3. It is less affected by extremely large values than does the AM.

4. It is capable of further algebraic manipulation.

5. It gives equal weight to all the values.

6. It is not much affected by fluctuations of sampling.

7. It is used in finding average of values which are in geometric progression.

8. It is the appropriate average for averaging the rates of change (e.g., the rate of

change in income, population, etc.) and ratios (e.g., price indices).

Disadvantages:

1. It is neither easy to calculate nor to understand.

2. It vanishes if any item in the data is zero.

3. It cannot be computed if any value is negative.

4. It may locate the value at a point at which few or none of the actual values lie.

Harmonic Mean:

Advantages:

1. HM is rigidly defined by a mathematical formula.

2. It is based on all the values.

3. It is capable of further algebraic manipulation.

4. It is not much affected by fluctuations of sampling.

5. It is an appropriate average for averaging time rates (e.g., speeds per hour) and

ratios (e.g., units purchased per rupee, etc.)

Disadvantages:

1. It is neither simple to understand nor easy to calculate.

2. The HM is greatly influenced by extremely small values.

3. It cannot be determined if any value in the data is zero.

Median:

Advantages:

1. The median is simple to understood and easy to calculate.

2. It is not affected by extremely large or extremely small values.

3. It can be computed from an open-end frequency table.

4. It is the most appropriate average in a highly skewed distribution, e.g., the

distribution of wages, incomes, etc.

5. It can be located even if the items are not capable of quantitative measurement.

For instance, we may arrange a number of pieces of blue cloth in order of

intensity of their colour and find the piece with the median colour.

6. It is not affected by changes in the values of the items.

7. The sum of absolute deviations (i.e., the ignoring negative signs) is the smallest

when measured from median than from any other average.

Disadvantages:

1. It is not rigidly defined.

2. It is not based on all the values.

3. It is not capable of further algebraic manipulation.

4. It is necessary to arrange the values in an array before finding the median, which

is a tedious work.

Mode:

Advantages:

1. It is simple to understand and easy to calculate. It can be located simply by

inspection in discrete distributions.

2. It is not influenced by extremely large or extremely small values.

3. It can be determined even in an open-end frequency table.

Disadvantages:

1. It is not well-defined. Sometimes, a distribution may have no mode at all or it

may have more than one mode.

2. It is not based on all the values.

3. It is not capable of further algebraic manipulation.

4. There will be no well-defined mode if the distribution consists of small number of

values.