A batsman scores runs in 10 innings: 38, 70, | vedclass

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(A) First,arrange the given data in ascending order:$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$Since the number of observations $n = 10$ (even),the media

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$8.6$

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(A) First,arrange the given data in ascending order:$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$Since the number of observations $n = 10$ (even),the median $M$ is the mean of the $5^{th}$ and $6^{th}$ terms:$M = \frac{46 + 48}{2} = 47$The mean deviation about the median is given by $\frac{1}{n} \sum |x_i - M|$.Calculating $|x_i - 47|$ for each term:$|34-47| = 13$$|38-47| = 9$$|42-47| = 5$$|44-47| = 3$$|46-47| = 1$$|48-47| = 1$$|54-47| = 7$$|55-47| = 8$$|63-47| = 16$$|70-47| = 23$Sum of absolute deviations = $13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86$Mean deviation = $\frac{86}{10} = 8.6$.

$x_i$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f_i$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

$A$ batsman scores runs in $10$ innings: $38, 70, 48, 34, 42, 55, 63, 46, 54, 44$. The mean deviation about the median is:

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