Question

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

$\begin{array}{ccccccccc}\text{Age}& 16-20& 21-25& 26-30& 31-35& 36-40& 41-45& 46-50& 51-55\\ \text{No. of persons}& 5& 6& 12& 14& 26& 12& 16& 9\end{array}$

Answer 1

Since the function is not continuos, we substract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains same, while the function becomes continuos.

Thus, the mean distribution table will be as follows :

$\begin{array}{cccccc}\text{Age}& \text{Number of persons}\left(f_i\right)& \text{Midpoint}{x}_i& \text{Cumulative Frequency}& |d_i|=|x_i-38|& f_i{d}_i\\ 15.5-20.5& 5& 18& 5& 20& 100\\ 20.5-25.5& 6& 23& 11& 15& 90\\ 25.5-30.5& 12& 28& 23& 10& 120\\ 30.5-35.5& 14& 33& 37& 5& 70\\ 35.5-40.5& 26& 38& 63& 0& 0\\ 40.5-45.5& 12& 43& 75& 5& 60\\ 45.5-50.5& 16& 48& 91& 10& 160\\ 50.5-55.5& 9& 53& 100& 15& 135\\ & N={\sum }_{i=1}^8{f}_i=100& & & & {\sum }_{i=1}^8{f}_i{d}_i=735\end{array}$

$N=100$

$\Rightarrow \frac{N}{2}=50$

Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5-40.5.

$\therefore$ l = 35.5, F = 37, f = 26, h = 5

Median = $l+\frac{\frac{N}{2}-F}{f}\times h$

$=35.5+\frac{(50-37)}{26}\times 5=35.5+2.5=38$

Mean deviation about the median age $=\frac{{\sum }_{i=1}^8{f}_i|d_i|}{N}=\frac{735}{100}=7.35$

Thus, the mean deviation from the median age is 7.35 years.