The mean deviation about the median of the di | vedclass

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(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$. Here,the number of terms $n = 8$,which is even. So,the median is the average of th

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$5$

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(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$. Here,the number of terms $n = 8$,which is even. So,the median is the average of the $4^{th}$ and $5^{th}$ terms. Median $= \frac{12 + 14}{2} = \frac{26}{2} = 13$. Now,we calculate the mean deviation about the median using the formula $\frac{\sum |x_i - 	ext{Median}|}{n}$. $x_i$$|x_i - 13|$$4$$9$$4$$9$$7$$6$$12$$1$$14$$1$$15$$2$$15$$2$$23$$10$ Sum of absolute deviations $\sum |d_i| = 9 + 9 + 6 + 1 + 1 + 2 + 2 + 10 = 40$. Mean deviation about median $= \frac{40}{8} = 5$.

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

The mean deviation about the median of the discrete data $12, 15, 7, 4, 4, 15, 23, 14$ is

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Class Interval $0$–$2$ $2$–$4$ $4$–$6$ $6$–$8$ $8$–$10$
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