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}$.

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

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

$x_i$	$5$	$7$	$9$	$10$	$12$	$15$
$f_i$	$8$	$6$	$2$	$2$	$2$	$6$

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

Similar Questions

Consider the following distribution:
$x_i$ $2$ $4$ $6$ $8$ $10$
$f_i$ $1$ $2$ $3$ $2$ $1$

The sum of the mean deviation from the mean and the mean deviation from the median of this distribution is:

Find the mean deviation about the median for the data.

${x_i}$ $5$ $7$ $9$ $10$ $12$ $15$

${f_i}$ $8$ $6$ $2$ $2$ $2$ $6$

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

The mean deviation about the mean for the data:

$x_i$ $5$ $7$ $9$ $10$ $12$ $15$

$f_i$ $8$ $6$ $2$ $2$ $2$ $6$

is equal to: (in /$13$)

If the mean deviation about the median for the numbers $3, 5, 7, 2k, 12, 16, 21, 24$ arranged in ascending order is $6$,then the median is:

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