Find the mean deviation about the mean for the following data.

Marks obtained	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
No of students	$2$	$3$	$8$	$14$	$8$	$3$	$2$

(D) We make the following table from the given data:

Marks obtained	Number of students $(f_i)$	Mid-points $(x_i)$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$10-20$	$2$	$15$	$30$	$30$	$60$
$20-30$	$3$	$25$	$75$	$20$	$60$
$30-40$	$8$	$35$	$280$	$10$	$80$
$40-50$	$14$	$45$	$630$	$0$	$0$
$50-60$	$8$	$55$	$440$	$10$	$80$
$60-70$	$3$	$65$	$195$	$20$	$60$
$70-80$	$2$	$75$	$150$	$30$	$60$
Total	$N=40$	-	$1800$	-	$400$

Here,$N = \sum f_i = 40$ and $\sum f_i x_i = 1800$.
The mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1800}{40} = 45$.
The mean deviation about the mean is given by $M.D.(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.
$M.D.(\bar{x}) = \frac{400}{40} = 10$.