If the mean deviation about the median of x, | vedclass

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Question

(A) The given observations are $x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x$. Since there are $n = 10$ observations,the median is the average of the $5^{th

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The given observations are $x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x$. Since there are $n = 10$ observations,the median is the average of the $5^{th}$ and $6^{th}$ terms: $	ext{Median} = \frac{5x + 6x}{2} = 5.5|x|$.The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{10} |x_i - 	ext{Median}| = 30$.Substituting the values: $\frac{1}{10} (|x - 5.5x| + |2x - 5.5x| + |3x - 5.5x| + |4x - 5.5x| + |5x - 5.5x| + |6x - 5.5x| + |7x - 5.5x| + |8x - 5.5x| + |9x - 5.5x| + |10x - 5.5x|) = 30$.This simplifies to: $\frac{1}{10} (| -4.5x | + | -3.5x | + | -2.5x | + | -1.5x | + | -0.5x | + | 0.5x | + | 1.5x | + | 2.5x | + | 3.5x | + | 4.5x |) = 30$.Since $|-a| = |a|$,we have: $\frac{2}{10} (4.5 + 3.5 + 2.5 + 1.5 + 0.5) |x| = 30$.$\frac{2}{10} (12.5) |x| = 30$.$2.5 |x| = 30$.$|x| = \frac{30}{2.5} = 12$.

$x_i$	$5$	$10$	$15$	$20$	$25$
$f_i$	$7$	$4$	$6$	$3$	$5$

$x$	$10$	$11$	$12$	$13$
$f$	$6$	$12$	$18$	$12$

If the mean deviation about the median of $x, 2x, 3x, 4x, 5x, 6x, 7x, 8x, 9x, 10x$ is $30$,then $|x|$ equals:

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