The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, \(\bar{x}=\frac{4+7+8+10+12+13+17}{8}=\frac{80}{8}=10\)
The deviations of the respective observations from the mean \(\bar{x},i.e.x_i-\bar{x},\) are:
6, 3, 2, 1, 0, 2, 3, 7
The absolute values of the deviations, i.e. \(|x_i-\bar{x}|\), are:
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
M.D.\(\bar{x}=\frac{\sum_{i=1}^{8}|x_i-\bar{x}|}{8}=\frac{6+3+2+1+0+2+3+7}{8}\)
\(=\frac{24}{8}=3\)