Question

Find the mean deviation from the mean and from median of the following distribution :

$\begin{array}{cccccc}\text{Marks}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \text{No. of students}& 5& 8& 15& 16& 6\end{array}$

Answer 1

Computation of M.D.from Median

$\begin{array}{ccccccccc}\text{Marks}& \text{Students}& x_i& \text{Cum.Freq.}& |d_i|=|x_i-28|& f_i{d}_i& f_i{x}_i& |x_i-27|& f_i|x_i-27|\\ 0-10& 5& 5& 5& 23& 115& 25& 22& 110\\ 10-20& 8& 15& 13& 13& 104& 120& 12& 96\\ 20-30& 15& 25& 28& 3& 45& 375& 2& 30\\ 30-40& 16& 35& 44& 7& 112& 560& 8& 128\\ 40-50& 6& 45& 50& 17& 102& 270& 18& 108\\ & & & & & {\sum }_{i=1}^5{f}_i{d}_i=478& & T=1350& {\sum }_{i=1}^8{f}_i|x_i-27|=472\end{array}$

$N=50,\frac{N}{2}=25$

The cumulative frequency just greater than $\frac{N}{2}=25$ is 28 and the corresponding class is 20 - 30

Thus the median class is 20 - 30

Using formula

$\therefore l=20,F=13,f=15,h=10$

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

Substituting the values

Median $=20+\frac{25-13}{15}\times 10=20+8=28$

Mean distribution on from the median $={\sum }_{i=1}^5\frac{f_i|d_i|}{N}=\frac{478}{50}=9.56$

Mean $\left(\overline{X}\right)={\sum }_{i=1}^5\frac{f_i{x}_i}{N}=\frac{1350}{50}=27$

M.D. from the mean $=\frac{1}{50}\times {\sum }_{i=1}^5{f}_i|x_i-27|=\frac{472}{50}=9.44$

M.D. from the mean are 9.56 and 9.44 respectively.