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

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq	$6$	$7$	$15$	$16$	$4$	$2$

(10.16) To calculate the mean deviation about the median,we first construct the frequency distribution table:

Class	Freq $(f_i)$	$c.f.$	Mid-point $(x_i)$	$|x_i - M|$	$f_i |x_i - M|$
$0-10$	$6$	$6$	$5$	$23$	$138$
$10-20$	$7$	$13$	$15$	$13$	$91$
$20-30$	$15$	$28$	$25$	$3$	$45$
$30-40$	$16$	$44$	$35$	$7$	$112$
$40-50$	$4$	$48$	$45$	$17$	$68$
$50-60$	$2$	$50$	$55$	$27$	$54$
Total	$N=50$	-	-	-	$508$

Here,$N = 50$,so $\frac{N}{2} = 25$. The cumulative frequency just greater than $25$ is $28$,which corresponds to the class $20-30$.
Thus,the median class is $20-30$.
Using the formula for median: $M = l + \frac{\frac{N}{2} - C}{f} \times h$
Where $l = 20, C = 13, f = 15, h = 10$.
$M = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{120}{15} = 20 + 8 = 28$.
Mean deviation about median $= \frac{1}{N} \sum f_i |x_i - M| = \frac{508}{50} = 10.16$.