Question

Calculate the mean deviation about median age for the age distribution of 100 persons given below

Age (in years)	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
Number	5	6	12	14	26	12	16	9

Answer 1

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age| Number $f_i$| Cumulative frequency (c.f.) | Mid point $x_i$| $|x_i-Med.|$| $f_i|x_i-Med.|$|
---|---|---|---|---|---
15.5-20.5| 5| 5| 18| 20| 100
20.5-25.5| 6| 11| 23| 15| 90
25.5-30.5| 12| 23| 28| 10| 120
30.5-35.5| 14| 37| 33| 5| 70
35.5-40.5| 26| 63| 38| 0| 0
40.5-45.5| 12| 75| 43| 5| 60
45.5-50.5| 16| 91| 48| 10| 160
50.5-55.5| 9| 100| 53| 15| 135
| 100| | | | 735

The class interval containing the $(\frac{N}{2})^{th}$ or 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5.is the median class.

It is known that,

Median= $I+\frac{\frac{N}{2}-c}{f}h$

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

$Median=35.5+\frac{50-37}{26}×5=35.5+\frac{13×5}{26}=35.5+2.5=38$

Thus, mean deviation about the median is given by,

$M.D.(\bar{x})=\frac{1}{N}\sum_{i=1}^{8}f_i|x_i-M|=\frac{1}{100}×735.1=7.35$