Question

Calculate the mean deviation about the median for the following data:

$\begin{array}{ccccccccc}\text{Class}& 16-20& 21-25& 26-30& 31-35& 36-40& 41-45& 46-50& 51-55\\ \text{Frequency}& 5& 6& 12& 14& 26& 12& 16& 9\end{array}$

Answer 1

Converting the given series into an exclusive series, we prepare the table, given below:

$\begin{array}{cccc}\text{Class}& \text{Frequency}\left(f_i\right)& cf& \text{Midpoint}\left(x_i\right)\\ 15.5-20.5& 5& 5& 18\\ 20.5-25.5& 6& 11& 23\\ 25.5-30.5& 12& 23& 28\\ 30.5-35.5& 14& 37& 33\\ 35.5-40.5& 26& 63& 38\\ 40.5-45.5& 12& 75& 43\\ 45.5-50.5& 16& 91& 48\\ 50.5-55.5& 9& 100& 53\\ & N=100& & \end{array}$

Thus N = 100 and therefore, $\frac{N}{2}=50$

$\Rightarrow$ median class is 35.5 - 40.5

$\Rightarrow$ L = 35.5, f = 26, h = 5 and c = 37.

$\therefore \text{median}=l+\frac{(\frac{N}{2}-c)}{f}\times h$

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

Thus, M = 38.

Now, we prepare the table given below.

$\begin{array}{cccc}{f}_i& x_i& |x_i-M|& f_i\times |x_i-M|\\ 5& 18& 20& 100\\ 6& 23& 15& 90\\ 12& 28& 10& 120\\ 14& 33& 5& 70\\ 26& 38& 0& 0\\ 12& 43& 5& 60\\ 16& 48& 10& 160\\ 9& 53& 15& 135\\ N=100& & & 735\end{array}$

Thus, $\sum f_i\times |x_i-M|=735$ and N = 100

$\therefore MD\left(M\right)=\frac{\sum f_i\times |x_i-M|}{N}=\frac{735}{100}=7.35$

Hence, the mean deviation about the median is 7.35