Find the mean deviation about the median for | vedclass

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(A) The given data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.Here,the number of observations is $n = 10$,which is even.Arranging the data in ascendi

Vedclass · Accepted Answer

$7.0$

Vedclass · Answer

(A) The given data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.Here,the number of observations is $n = 10$,which is even.Arranging the data in ascending order,we obtain:$36, 42, 45, 46, 46, 49, 51, 53, 60, 72$.Median $M = \frac{(\frac{n}{2})^{	ext{th}} 	ext{ observation} + (\frac{n}{2} + 1)^{	ext{th}} 	ext{ observation}}{2}$.$M = \frac{5^{	ext{th}} 	ext{ observation} + 6^{	ext{th}} 	ext{ observation}}{2} = \frac{46 + 49}{2} = \frac{95}{2} = 47.5$.The absolute deviations $|x_i - M|$ are:$|36 - 47.5| = 11.5$$|42 - 47.5| = 5.5$$|45 - 47.5| = 2.5$$|46 - 47.5| = 1.5$$|46 - 47.5| = 1.5$$|49 - 47.5| = 1.5$$|51 - 47.5| = 3.5$$|53 - 47.5| = 5.5$$|60 - 47.5| = 12.5$$|72 - 47.5| = 24.5$The mean deviation about the median is $\frac{\sum |x_i - M|}{n} = \frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10} = \frac{70}{10} = 7.0$.

Find the mean deviation about the median for the data $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.

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