# Statistics - Continuous Series Arithmetic Median

When data is given based on ranges along with their frequencies. Following is an example of continous series −

 Items Frequency 0-5 5-10 10-20 20-30 30-40 2 5 1 3 12

## Formula

$Median = {L} + \frac{(\frac{n}{2}\ -\ c.f.)}{f} \times {i}$

Where −

• ${L}$ = Lower limit of median class, median class is that class where $\frac{n}{2}^{th}$ item is lying.

• ${c.f.}$ = Cumulative frequency of the class preceding the median class.

• ${f}$ = Frequency of median class.

• ${i}$ = Class interval of median class.

Arithmetic Median is a useful measure of central tendency in case the data type is nominal data. Since it is a positional average, it does not get affected by extreme values.

## Example

Problem Statement

In a study conducted in an organization, the distribution of income across the workers is observed. Find the the median wage of the workers of the organization.

• 06 men get less than Rs. 500

• 13 men get less than Rs. 1000

• 22 men get less than Rs. 1500

• 30 men get less than Rs. 2000

• 34 men get less than Rs. 2500

• 40 men get less than Rs. 3000

Solution

Given are the cumulative frequencies of the workers. Hence we first find the simple frequency and present the data in tabular form.

Income

(rs.)

M.P.

m

Frequency

f

(m-1250)/500

d

fd

c.f

0 - 500 250 6 -2 -12 6
500 - 1000 750 7 -1 -7 13
1000 - 1500 1250 9 0 0 22
1500 - 2000 1750 8 1 8 30
2000 - 2500 2250 4 2 8 34
2500 - 3000 2750 6 3 18 40
N = 40   ∑ fd = 15

In order to simplify the calculation, a common factor i = 500 has been taken. Using the following formula for calculating median wage.

$Median = {L} + \frac{(\frac{n}{2}\ -\ c.f.)}{f} \times {i}$

Where −

• ${L}$ = 1000

• $\frac{n}{2}$ = 20

• ${c.f.}$ = 13

• ${f}$ = 9

• ${i}$ = 500

Thus

$Median = {1000} + \frac{(20\ -\ 13)}{9} \times {500} \\[7pt] \, = {1000 + 388.9} \\[7pt] \, = {1388.9}$

As 1388.9 ≃ 1389.

The median wage is Rs. 1389.