The mean deviation about the mean for the dat | vedclass

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Question

(C) First, calculate the total frequency $N = \sum f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26$.
Next, calculate the sum of the products $\sum f_i x_i = (5 	im

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) First, calculate the total frequency $N = \sum f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26$.
Next, calculate the sum of the products $\sum f_i x_i = (5 	imes 8) + (7 	imes 6) + (9 	imes 2) + (10 	imes 2) + (12 	imes 2) + (15 	imes 6) = 40 + 42 + 18 + 20 + 24 + 90 = 234$.
The mean $\bar{x}$ is given by $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{234}{26} = 9$.
The mean deviation about the mean is calculated as $MD = \frac{\sum f_i |x_i - \bar{x}|}{N}$.
$MD = \frac{8|5-9| + 6|7-9| + 2|9-9| + 2|10-9| + 2|12-9| + 6|15-9|}{26}$.
$MD = \frac{8(4) + 6(2) + 2(0) + 2(1) + 2(3) + 6(6)}{26} = \frac{32 + 12 + 0 + 2 + 6 + 36}{26} = \frac{88}{26} = \frac{44}{13}$.

$x_i$	$5$	$7$	$9$	$10$	$12$	$15$
$f_i$	$8$	$6$	$2$	$2$	$2$	$6$

$x_i$	$10$	$30$	$50$	$70$	$90$
$f_i$	$4$	$24$	$28$	$16$	$8$

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

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

The mean deviation about the mean for the data:

$x_i$ $5$ $7$ $9$ $10$ $12$ $15$

$f_i$ $8$ $6$ $2$ $2$ $2$ $6$

is equal to: (in /$13$)

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${x_i}$ $5$ $7$ $9$ $10$ $12$ $15$

${f_i}$ $8$ $6$ $2$ $2$ $2$ $6$

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 data: $x_i$ $5$ $7$ $9$ $10$ $12$ $15$ $f_i$ $8$ $6$ $2$ $2$ $2$ $6$ is equal to: (in /$13$)