If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged i…

Question

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

Accepted Answer

Median $$ = {{2k + 12} \over 2} = k + 6$$

Mean deviation $$ = \sum {{{|{x_i} - M|} \over n} = 6} $$

$$ \Rightarrow {{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)} \over 8}$$

$$\therefore$$ $${{58 - 2k} \over 8} = 6$$

$k = 5$

Median $$ = {{2 \times 5 + 12} \over 2} = 11$$

Solutions · 1