Prof. Rajkumar Teotia

Institute of Advanced Management and Research (IAMR)

Address: 9th Km Stone, NH-58, Delhi-Meerut Road, Duhai,Ghaziabad (U.P) - 201206

Ph:0120-2675904/905 Mob:9999052997 Fax: 0120-2679145

e mail: rajkumarteotia@iamrindia.com

Median is the central value of the variable that divide the series

into two equal parts in such a way that half of the items lie

above the value and the remaining half lie below this value.

Median is defined as the value of the middle item (or the mean

of the values of the two middle items) when the data are

arranged in an ascending or descending order of magnitude.

Thus, in an ungrouped frequency distribution if the n values are

arranged in ascending or descending order of magnitude, the

median is the middle value if n is odd.

When n is even, the median is the mean of the two middle

values.

Example

Suppose we have the following series:

15, 19,21,7, 10,33,25,18 and 5

Solution

We have to first arrange it in either ascending or descending

order.

These figures are arranged in an ascending order as follows:

5,7,10,15,18,19,21,25,33

Now as the series consists of odd number of items, to find out the

value of the middle item, we use the formula

Where n is the number of items

In this case, n is 9, as such

N + 1 = 9 + 1 = 5

2 2

That is, the size of the 5th item is the median.

So median is 18

Example

Suppose we have the following series:

15, 19,21,7, 10,33,25,18, 5 and 23

Solution

We have to first arrange it in either ascending or descendingorder. These figures are arranged in an ascending order asfollows:

5, 7, 10, 15, 18, 19, 21,23,25,33

Now as the series consists of even number of items, to find outthe value of the middle item, we use the formula

Where n is the number of items

In this case, n is 10, as such

N + 1 = 10 + 1 = 5.5

2 2

That is, the size of the 5.5th item is the median. We have to take

the average of the values of 5th and 6th item. This means an

average of 18 and 19, which gives the median as 18.5.

In the case of a grouped series, the median is calculated with

the help of the following formula:

Where,

L = Lower limit of median class

P.c.f = Previous commutative frequency of median class

f = frequency of median class.

i = Size of the median class.

N = total no of observation or the total of the frequency.

Median = L + N - P.c.f x i

2

f

Example – From the following data, calculate median.

Solution-

Step I- First we will find out the commutative frequency

Marks 0-10 10-20 20-30 30-40 40-50 50-60

No. of students 10 20 30 50 40 30

Marks(x) No. of students (f) Commutative

frequency

C.f

0-10

10-20

20-30

30-40

40-50

50-60

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Step II - Size of N item = size of 180 item = 90th item

2 2

Step III-

Commutative frequency which includes 90th = 110 Class

corresponding to 110 = 30 – 40 (is the median class)

Marks(x) No. of students

(f)

Commutative

frequency

C.f

0-10

10-20

20-30

30-40

40-50

50-60

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Median Class

PCF

fL

Step iv – now we will apply the following formula

Median = 30 + 90 – 60 x 10

50

Median = 36

Median = L + N - P.c.f x i

2

f

Example: from the following data calculate median

Marks 45 55 25 35 5 15

No. of students 40 30 30 50 10 20

Solution-

Step I- First we will find out the commutative frequency

Marks in

ascending

order (x)

No. of students (f) Commutative

frequency

C.f

5

15

25

35

45

55

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Step II - Size of N + 1 item = size of 181 item = 90.5th item

2 2

Step III- Commutative frequency which includes 90.5th = 110

Median = size of item corresponding to 110 = 35

Unlike the arithmetic mean, the median can be computedfrom open-ended distributions. This is because it is locatedin the median class-interval, which would not be an open-ended class

As it is not influenced by the extreme values, it is preferredin case of a distribution having extreme values.

In case of the qualitative data where the items are notcounted or measured but are scored or ranked, it is the mostappropriate measure of central tendency.

The values of a variate that divide the series or the distribution into four

equal parts are known as quartiles. Since three points are required to

divide the data into four equal parts, we have three quartiles Q1, Q2, Q3.

The first quartile (Q1):- it is known as a lower quartile, is the value of a

variate below which there are 25% of the observation and above which

there are 75% of the observations.

The second quartile (Q2):- it is known as a middle quartile or median, is

the value of a variate which divides the distribution into two equal parts.

It means there are 50% of the observations above it and 50% of the

observations below it.

The Third quartile (Q3):- it is known as an upper quartile, is the value

of a variate below which there are 75% of the observations and above

which there are 25% of the observations.

Q1 = size of N + 1 th item

4

Q2 = size of 2( N + 1) th item

4

Q3 = size of 3( N + 1) th item

4

Example: - from the following data calculate first and third quartile.

Solution:-

Step I: - Calculation of commutative Frequencies

Marks 5 15 25 35 45 55

No. of students 10 20 30 50 40 30

Marks No. of students (f) Commutative

frequency

C.f

5

15

25

35

45

55

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Step II: -

Q1 = size of N + 1 th item

4

= Size of 180 + 1 th item = 181 = 45.25 th item

4 4

= size of 45.25 th item = 25, So Q1 = 25

Step III: -

Q3 = size of 3(N + 1) th item

4

Q3 = 135.7 th item

Q3 = 45

Computation of Quartiles for grouped data:-

Q1 = L + N - P.c.f x i

4

f

Q2 = L + N - P.c.f x i

2

f

Q3 = L + 3N - P.c.f x i

4

f

DECILES

The deciles of a variate that divide the series or the distribution

into ten equal parts are called deciles. Each part contains 10% of

the total observations. Since nine points are required to divide the

data into 10 equal parts, we have nine deciles that are D1 to D9

Computation of Deciles for ungrouped data and

discrete series(after arranging the size of item in

ascending or descending order):-

DJ = size of J (N + 1) th item

10

Where,

J = 1, 2, 3, 4, 5, 6, 7, 8, 9,

Example: - from the following data calculate first and second Deciles.

Solution:-

Step I: - Calculation of commutative Frequencies

Marks 5 15 25 35 45 55

No. of students 10 20 30 50 40 30

Marks No. of students (f) Commutative frequency

C.f

5

15

25

35

45

55

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Step II: -

D1 = size of N + 1 th item

10

= Size of 180 + 1 th item = 181 = 18.1 th item

10 10

= size of 18.1 th item = 25, So D1 = 15

Step III: -

D2 = size of 2(N + 1) th item

10

D2 = 36.2 th item

D2 = 25

Where,

J = 1, 2, 3, 4, 5, 6, 7, 8, 9,

DJ = L + J N - P.c.f x i

10

f

The value of a variate which divides a given series or

distribution into 100 equal parts are known as percentiles.

Each percentile contains 1% of the total number of

observations. Since ninety nine points are required to divide

the data into 100 equal parts, we have 99 percentiles, P1 to P100

PJ = size of J (N + 1) th item

100

Where, J = 1 to 100

Computation of Percentiles for grouped data:-

Where, J = 1 to 100

PJ = L + J N - P.c.f x i

100

f

Example: - from the following data calculate Q1, D8 and P10.

Solution:-

Step I: -Calculation of commutative Frequencies

Marks 0-10 10-20 20-30 30-40 40-50 50-60

No. of students 10 20 30 50 40 30

Marks No. of students

(f)

Commutative

frequency

C.f

0-10

10-20

20-30

30-40

40-50

50-60

10

20

30

50

40

30

10

30

60

110

150

180

N = 180

Q1

D8

P10

Step II: - Calculation of Q1

N th item = 180 = 45th item

4 4

Q1 = 20 + 45 – 30 x 10 = 25

30

Q1 = L + N - P.c.f x i

4

f

Step III: - Calculation of D8

8 N th item = 8 x 180 = 144th item

10 10

D8 = 40 + 144 – 110 x 10 = 48.5

40

D8 = L + 8N - P.c.f x i

10

f

Step IV: - Calculation of P10

10 N th item = 10 x 180 = 18th item

100 100

P10 = 10 + 18 – 10 x 10 = 14

20

P10 = L + 10N - P.c.f x i

100

f

Mode is often said to be that value in a series which occurs most

frequently or which has the greatest frequency. But it is not

exactly true for every frequency distribution. Rather it is that

value around which the items tend to concentrate most heavily.

Calculation of mode in case of ungrouped data

Example- Find the mode of the following series: 8, 9, 11, 15, 16,

12, 15, 3, 7, 15

Solution- There are ten observations in the series wherein the

figure 15 occurs maximum number of times three. The

mode is therefore 15.

In the case of grouped data, mode is determined by the following formula:

Where,

MO = Mode.

L = Lower limit of the modal class.

∆1 = The difference between the frequency of the modal class and the frequency of the pre modal class.

∆2 = The difference between the frequency of the modal class and the frequency of the post-modal class.

i = The size of the modal class

MO = L + ∆1 x i

∆1 + ∆2

Example- calculate the modal sales of the 100 companies from the following data

Solution-

Since the maximum frequency is 30 is in the class 64-66, therefore 64-66 is the

modal class

Sales in Rs(lakhs) 58-60 60-62 62-64 64-66 66-68 68-70 70-72

No. of companies 12 18 25 30 10 3 2

Sales in

Rs(lakhs)

No. of companies

58-60 12

60-62 18

62-64 25

64-66 30

66-68 10

68-70 3

70-72 2

Modal class

Mode is determined by the following formula

MO = 64 + 5 x 2

5 + 20

MO = 64.4

MO = L + ∆1 x i

∆1 + ∆2

Example: -

from the following data, calculate Mode:

Marks 5 10 11 12 13 14 15 16 18 20

No. of students 4 6 5 10 20 22 24 6 2 1

Solution:-

First we will do grouping of the above data with the help of

grouping table. A grouping table must have six columns.

Marks Column 1 Column 2 Column 3 Column 4 Column 5 Column 6

5 4 x x x

10 6 10 15 x

11 5 11 21

12 10 15 35

13 20 30

14 22

15

16 6 30 32

18 2 8 9 x

20 1 3 x x x

24

42

46

52

66

52

ANALYSIS TABLE

The highest total in the analysis table is five. The item

corresponding to it is 14. Therefore, the mode is 14

Column No. Marks

5 10 11 12 13 14 15 16 18 20

1 1

2 1 1

3 1 1

4 1 1 1

5 1 1 1

6 1 1 1

TOTAL 1 3 5 4 1

Mode can be also determined indirectly by applying the

following formula:

Merits of Mode

In many cases it can be found by inspection.

It is not affected by extreme values.

It can be calculated for distributions with open end classes.

It can be located graphically.

It can be used for qualitative data.

Mode = 3 median - 2 mean

Demerits of Mode

It is not based on all values.

It is not capable of further mathematical treatment.

It is much affected by sampling fluctuations.

Having discussed mean, median and mode, we now turn to the

relationship amongst these three measures of central tendency. We shall

discuss the relationship assuming that there is a unimodal frequency

distribution.

1-Symmetrical Distribution

When a distribution is symmetrical, the mean, median and mode are the

same as is shown below in the following figure.

Mean = median = mode

2-Asymmetrical Distribution

Asymmetrical distributions are of following type

Positively skewed

Negatively skewed

‘L’ shaped positively skewed

‘J’ shaped negatively skewed

Positively skewed

Mean ˃ Median ˃ Mode

Negatively skewed

Mean ˂ Median ˂ Mode

‘L’ Shaped Positively skewed

Mean ˂ Mode

Mean ˂ Median

‘J’ Shaped Negatively Skewed

Mean ˃ Mode

Mean ˃ Median

Moderately Skewed /Asymmetrical Distribution

Under Peak of curve Divides area in

halves

Centre of Gravity

Mode Median Mean