Question

Find the mean deviation about the median for the data $13,17,16,14,11,13,10,16,11,18,12,17$.

Answer 1

Arranging the data in ascending order
$10,11,11,12,13,13,14,16,16,17,17,18$
Here the numbers of observations $n=12$ (even)
$M=\frac{{\left(\frac{12}{2}\right)}^{th}observation+{(\frac{12}{2}+1)}^{th}observation}{2}$
$M=\frac{6^{th}observation+7^{th}observation}{2}$
$M=\frac{13+14}{2}=\frac{27}{2}=13.5$
The deviations of the respective observation from the median, i.e.$x_i-M$ are

$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,0.5,2.5,2.5,3.5,3.5,4.5$

The absolute values of the deviations $\mid x_i-M\mid$ are $3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5$
The required mean deviation about the median is
$M.D.\left(M\right)=\frac{{\sum }_{i=1}^{12}|x_1-M|}{12}$

$=\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}$

$=\frac{28}{12}=2.33$