Question

Find the mean deviation about median for the following data.

$CI$	$8-12$	$13-17$	$18-22$	$23-27$	$28-32$	$33-37$	$38-42$	$43-47$
$f$	$14$	$8$	$20$	$7$	$11$	$10$	$24$	$6$

Answer 1

$CI$	$f_i$	$cf$	$Mid-value$ $x_i$	$|x_i-M|$	$f_i|x_i-M|$
$7.5-12.5$	$14$	$14$	$10$	$18.21$	$254.94$
$12.5-17.5$	$8$	$22$	$15$	$13.21$	$105.68$
$17.5-22.5$	$20$	$42$	$20$	$8.21$	$164.2$
$22.5-27.5$	$7$	$49$	$25$	$3.21$	$22.47$
$27.5-32.5$	$11$	$60$	$30$	$1.79$	$19.69$
$32.5-37.5$	$10$	$70$	$35$	$6.79$	$67.9$
$37.5-42.5$	$24$	$94$	$40$	$11.79$	$282.96$
$42.5-47.5$	$6$	$100$	$45$	$16.79$	$100.74$
	$\sum f_i=100$				$\sum f_i|x_i-M|=3243.11$

$\Rightarrow$ Here, $N=100$, $\frac{N}{2}=\frac{100}{2}=50$

$\therefore$ Median class $=22.5-27.5$

$\Rightarrow$ $L=22.5,cf=42,f=7$ and $h=27.5-22.5=5$

$\Rightarrow$ $Median\left(M\right)=L+\frac{\frac{N}{2}-cf}{f}\times h$

$=22.5+\frac{50-42}{7}\times 5$

$=22.5+5.71$

$=28.21$

$\Rightarrow$ Mean deviation about the median $=\frac{\sum f_i|x_i-M|}{\sum f_i}$

$=\frac{3243.11}{100}$

$=32.43$