”;
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
”;