If the mean deviation about median for the nu | vedclass

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Question

Median $=\frac{2 k+12}{2}=k+6$

Mean deviation $=\sum \frac{\left|x_{i}-M\right|}{n}=6$

$(k+3)+(k+1)+(k-1)+(6-k)+(6-k)$

$\frac{+(10-k)+(15-k)+

Vedclass · Accepted Answer

$11$

Vedclass · Answer

Median $=\frac{2 k+12}{2}=k+6$

Mean deviation $=\sum \frac{\left|x_{i}-Might|}{n}=6$

$(k+3)+(k+1)+(k-1)+(6-k)+(6-k)$

$\frac{+(10-k)+(15-k)+(18-k)}{8}$

$	herefore \quad \frac{58-2 k}{8}=6$

$k=5$

Median $=\frac{2 	imes 5+12}{2}=11$

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

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