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(D) First,we calculate the mean $(\bar{x})$ of the given data.The midpoints $(x_i)$ of the intervals are $5, 15, 25, 35, 45$.The frequencies $(f_i)$ a

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$9.6$

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(D) First,we calculate the mean $(\bar{x})$ of the given data.The midpoints $(x_i)$ of the intervals are $5, 15, 25, 35, 45$.The frequencies $(f_i)$ are $6, 8, 10, 4, 2$.Total frequency $N = \sum f_i = 6 + 8 + 10 + 4 + 2 = 30$.Sum of products $\sum f_i x_i = (5 	imes 6) + (15 	imes 8) + (25 	imes 10) + (35 	imes 4) + (45 	imes 2) = 30 + 120 + 250 + 140 + 90 = 630$.Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{630}{30} = 21$.Now,calculate the mean deviation about the mean using the formula $	ext{M.D.}(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.$\sum f_i |x_i - 21| = 6|5-21| + 8|15-21| + 10|25-21| + 4|35-21| + 2|45-21|$$= 6(16) + 8(6) + 10(4) + 4(14) + 2(24) = 96 + 48 + 40 + 56 + 48 = 288$.$	ext{Mean deviation} = \frac{288}{30} = 9.6$.

$\text{Marks obtained}$	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$\text{Number of Boys}$	$6$	$8$	$10$	$4$	$2$

Class Interval	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$
Frequency	$1$	$2$	$3$	$2$	$1$

The mean deviation about the mean for the following data is:
$\text{Marks obtained}$ $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
$\text{Number of Boys}$ $6$ $8$ $10$ $4$ $2$

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The mean deviation about the mean for the following data is:$\text{Marks obtained}$$0-10$$10-20$$20-30$$30-40$$40-50$$\text{Number of Boys}$$6$$8$$10$$4$$2$