Question

The mean deviation about the median for the following data

$\begin{array}{ccccccc}\text{Marks}& 0-10& 10-20& 20-30& 30-40& 40-50& 50-60\\ \text{Number of students}& 6& 8& 14& 16& 4& 2\end{array}$

is

• A. $9.35$
• B. $10.34$
• C. $11.35$
• D. $12.33$

Answer 1

The correct option is B $10.34$

$\begin{array}{cccccc}\text{Class}& \text{Mid points}\left(x_i\right)& f_i& c.f.& |x_i-M|& f_i|x_i-M|\\ 0-10& 5& 6& 6& 22.85& 137.1\\ 10-20& 15& 8& 14& 12.85& 102.8\\ 20-30& 25& 14& 28& 2.85& 39.9\\ 30-40& 35& 16& 44& 7.15& 114.4\\ 40-50& 45& 4& 48& 17.15& 68.6\\ 50-60& 55& 2& 50& 27.15& 54.3\end{array}$

The class interval containing ${\frac{N}{2}}^{th}$ or ${25}^{th}$ term is $20-30$. Therefore $20-30$ is the median class.

Median $=l+\frac{\frac{N}{2}-f_1}{f}\times h$
$=20+\frac{25-14}{14}\times 10$
$=20+\frac{11}{14}\times 10$
$=20+7.85=27.85$

Mean deviation about median is given by $\frac{1}{N}\sum _1^6{f}_i|x_i-M|$
$=\frac{1}{50}\times 517.1=10.34$