Find the mean deviation about the mean for the following data :

We have, 39, 72, 48, 41, 43, 55, 60, 45, 54, 43

Mean of the given data is

$\overline{\mathrm{x}}=\frac{39+72+48+41+43+55+60+45+54+43}{10}=\frac{500}{10}=50$

The respective absolute values of the deviations from mean, i.e' $\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|$ are

$11,22,2,9,7,5,10,5,4,7$

Thus, the required mean deviation about the mean is

M. D. $(\overline{\mathrm{x}})=\frac{\sum_{\mathrm{i}=1}^{10}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{10}$

$=\frac{11+22+2+9+7+5+10+5+4+7}{10}=\frac{82}{10}=8.2$