Statistics - Continuous Series Arithmetic Median

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

Items 0-5 5-10 10-20 20-30 30-40
Frequency 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.

Calculator

Advertisements