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

Question

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 (A) 11.5 (B) 10.5 (C) 12 (D) 11

Tardigrade Admin · Accepted Answer

3,5,7,2 k , 12,16,21,24 Median =(2 k +12/2)= k +6 M.D. =( Sigma| x i - M |/8)=6 =( k +3)+( k +1)+( k -1)+(6- k )+(6- k )+(10- k )+(15- k )+(18- k )=48 =58-2 k =48 k =5 Median = k +6=11

Q. 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

$11.5$

$10.5$

12

11

$3, 5, 7, 2 k, 12, 16, 21, 24$
Median $= \frac{2 k + 12}{2} = k + 6$
M.D. $= \frac{Σ ∣ x _{i} - M ∣}{8} = 6$
$= (k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 -$
$k) + (10 - k) + (15 - k) + (18 - k) = 48$
$= 58 - 2 k = 48$
$k = 5$
Median $= k + 6 = 11$

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

11.5

10.5

12

11

3,5,7,2k,12,16,21,24 Median =22k+12​=k+6 M.D. =8Σ∣xi​−M∣​=6 =(k+3)+(k+1)+(k−1)+(6−k)+(6− k)+(10−k)+(15−k)+(18−k)=48 =58−2k=48 k=5 Median =k+6=11

Q. 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

$11.5$

$10.5$

$3, 5, 7, 2 k, 12, 16, 21, 24$
Median $= \frac{2 k + 12}{2} = k + 6$
M.D. $= \frac{Σ ∣ x _{i} - M ∣}{8} = 6$
$= (k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 -$
$k) + (10 - k) + (15 - k) + (18 - k) = 48$
$= 58 - 2 k = 48$
$k = 5$
Median $= k + 6 = 11$