Question

The mean deviation for n observations $x_1,x_2,...,x_n$ from their mean $\overline{X}$ is given by

(a) $\sum _{i=1}^n\left(x_i-\overline{X}\right)$ (b) $\frac{1}{n}\sum _{i=1}^n\left(x_i-\overline{X}\right)$ (c) $\sum _{i=1}^n{\left(x_i-\overline{X}\right)}^2$ (d) $\frac{1}{n}\sum _{i=1}^n{\left(x_i-\overline{X}\right)}^2$

Answer 1

The mean deviation for n observations $x_1,x_2,...,x_n$ from their mean $\overline{X}$ is $\frac{1}{n}\sum _{i=1}^n\left|x_i-\overline{X}\right|$.


Disclaimer: There is some printing error in option (b) given in the question. The answer would be option (b) if it given as $\frac{1}{n}\sum _{i=1}^n\left|x_i-\overline{X}\right|$.