Find the mean deviation about the median for the data 
$36,72,46,42,60,45,53,46,51,49$

Arranging the data in ascending order, we get
$36,42,45,46,46,49,51,53,60,72$
Here the number of observations is $10$, which is even. 
Median $M=\frac{{\left(\frac{10}{2}\right)}^{th}observation+{\left(\frac{10}{2}+1\right)}^{th}observation}{2}$
$=\frac{5^{th}observation+6^{th}observation}{2}$
$=\frac{46+49}{2}=\frac{95}{2}=47.5$
The deviations of the respective observations from the median i.e. $x_i-M$ are
 $-11.5,-5.5,-2.5,-1.5,-1.5,1.5,3.5,5.5,12.5,24.5$
The  absolute values of the deviations$\mid x_i-M\mid$ are 
$11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5$
Thus the required mean deviation about the median is 
$M.D.(M)=\frac{{\sum }_{i=1}^{10}\left|x_{}-M\right|}{10}$ = $\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$
$=\frac{70}{10}=7$