Question

Calculate mean deviation about median for following readings
$\begin{array}{ccccc}Class& 20-40& 40-60& 60-80& 80-100\\ F_i& 20& 44& 30& 40\end{array}$

• A. 18.6
• B. 18.7
• C. 18.8
• D. 18.9

Answer 1

The correct option is C

18.8

We need to calculate the median first.

$\begin{array}{ccccc}Class& 20-40& 40-60& 60-80& 80-100\\ f_i& 20& 44& 30& 40\\ c.f_i& 20& 64& 94& 134\end{array}$

Median$=l+\frac{\frac{N}{2}-c}{f}Xh$

Median class$=(60-80)$ class as it contains $(\frac{134}{2}{)}^{th}$ term

$\therefore l=60,N=134,f=30,h=20$.

$\therefore$ Median $=60+\left(\frac{\frac{134}{2}-64}{30}\right)20$

$=60+\left(\frac{67-64}{30}\right)20=60+2=62$

$\therefore M=62$

$\begin{array}{ccccc}Class& 20-40& 40-60& 60-80& 80-100\\ f_i& 20& 44& 30& 40\\ Meandeviations& |30-62|& |50-62|& |70-62|& |90-62|\\ & =32& =12& =8& =28\end{array}$

Mean deviation about median (M) $=\sum _{i-1}^4\frac{{\sum }_{i=1}^{4}fi|x_i-M|}{{\sum }_{0-1}^{4}fi}$

$=\frac{20\left(32\right)+44\left(12\right)+30\left(8\right)+40\left(28\right)}{134}$

$=\frac{640+528+240+1120}{134}$

$=2528134=18.8$