The mean deviation about the median of the nu | vedclass

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(C) The given numbers are $3x, 6x, 9x, \ldots, 81x$. This is an arithmetic progression with $n = 27$ terms. Since $n = 27$ is odd,the median is the $\

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(C) The given numbers are $3x, 6x, 9x, \ldots, 81x$. This is an arithmetic progression with $n = 27$ terms. Since $n = 27$ is odd,the median is the $\frac{n+1}{2}$-th term,which is the $14$-th term. The $14$-th term is $3x 	imes 14 = 42x$. The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - 	ext{Median}|$. Here,$	ext{MD} = \frac{1}{27} \sum_{k=1}^{27} |3kx - 42x| = \frac{3|x|}{27} \sum_{k=1}^{27} |k - 14| = \frac{|x|}{9} [\sum_{k=1}^{13} (14-k) + \sum_{k=15}^{27} (k-14)]$. Calculating the sums: $\sum_{k=1}^{13} (14-k) = 13+12+\ldots+1 = \frac{13 	imes 14}{2} = 91$. Similarly,$\sum_{k=15}^{27} (k-14) = 1+2+\ldots+13 = 91$. So,$	ext{MD} = \frac{|x|}{9} (91 + 91) = \frac{|x|}{9} 	imes 182 = 91$. Thus,$\frac{|x|}{9} 	imes 2 = 1 \implies |x| = \frac{9}{2}$.

$x_i$	$2$	$4$	$6$	$8$	$10$
$f_i$	$1$	$2$	$3$	$2$	$1$

The mean deviation about the median of the numbers $3x, 6x, 9x, \ldots, 81x$ is $91$. Then $|x| = $

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