Statistics - Arithmetic Median of Continous Series
When data is given based on ranges alongwith 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
As 1388.9 ≃ 1389.
The median wage is Rs. 1389.
