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(B) First,we arrange the observations in ascending order:$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$Here,$n = 10$ (even).The median $(M) = \frac{(\frac{n

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$8.6$

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(B) First,we arrange the observations in ascending order:$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$Here,$n = 10$ (even).The median $(M) = \frac{(\frac{n}{2})^{th} 	ext{ term} + (\frac{n}{2} + 1)^{th} 	ext{ term}}{2} = \frac{46 + 48}{2} = 47$.Now,calculate the sum of absolute deviations from the median:$\Sigma |x_i - M| = |34-47| + |38-47| + |42-47| + |44-47| + |46-47| + |48-47| + |54-47| + |55-47| + |63-47| + |70-47|$$= 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86$.Mean deviation about the median $= \frac{\Sigma |x_i - M|}{n} = \frac{86}{10} = 8.6$.

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

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

The mean deviation about the median for the observations $70, 42, 63, 34, 44, 54, 55, 46, 38, 48$ is:

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

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

${x_i}$ $2$ $5$ $6$ $8$ $10$ $12$

${f_i}$ $2$ $8$ $10$ $7$ $8$ $5$

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