If the mean deviation about the median for th | vedclass

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(D) The given data is $3, 5, 7, 2k, 12, 16, 21, 24$. The number of observations $n = 8$.Since $n$ is even,the median $M$ is the average of the $4^{th}

Vedclass · Accepted Answer

$11$

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(D) The given data is $3, 5, 7, 2k, 12, 16, 21, 24$. The number of observations $n = 8$.Since $n$ is even,the median $M$ is the average of the $4^{th}$ and $5^{th}$ terms: $M = \frac{2k + 12}{2} = k + 6$.Mean deviation about median is given by $\frac{1}{n} \sum |x_i - M| = 6$.Substituting the values: $\frac{|3-(k+6)| + |5-(k+6)| + |7-(k+6)| + |2k-(k+6)| + |12-(k+6)| + |16-(k+6)| + |21-(k+6)| + |24-(k+6)|}{8} = 6$.This simplifies to: $\frac{|-k-3| + |-k-1| + |-k+1| + |k-6| + |6-k| + |10-k| + |15-k| + |18-k|}{8} = 6$.Assuming $k$ is such that the terms maintain the order,the sum is $(k+3) + (k+1) + (k-1) + (6-k) + (6-k) + (10-k) + (15-k) + (18-k) = 58 - 2k$.$\frac{58 - 2k}{8} = 6 \implies 58 - 2k = 48 \implies 2k = 10 \implies k = 5$.Thus,the median $M = k + 6 = 5 + 6 = 11$.

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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