Find the mean deviation about the mean for the data 
$38,70,48,40,42,55,63,46,54,44$

The given data is $38,70,48,40,42,55,63,46,54,44$

Mean of the given data 
$\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}=50$

The deviations of the respective observation from the mean $\bar{x},i.e.x_i=\bar{x}$ are $-12,20,-2,-10,-8,5,13,-4,4,-6$.

The absolute values of the deviations i.e.$\mid x_1-\bar{x}\mid$ are $12,20,2,10,8,5,13,4,4,6$

The required mean deviation about the mean is 
$M.D.=\frac{{\sum }_{i=1}^{10}\mid x_i-\bar{x}\mid }{10}$
$=\frac{12+10+2+10+8+5+13+4+4+6}{10}$
$=\frac{84}{10}$
$=8.4$