Find the mean deviation about the mean for the data

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 observations 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. $\left|x_{i}-\bar{x}\right|$, are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

$\operatorname{M.D} \cdot(\bar{x})=\frac{\sum_{i=1}^{10}\left|x_{t}-\bar{x}\right|}{10}$

$=\frac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\frac{84}{10}$

$=8.4$