Question

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

• A. ${\sum }_{i=1}^n(x_i-\overline{X})$
• B. $\frac{1}{n}{\sum }_{i=1}^n(x_i-\overline{X})$
• C. ${\sum }_{i=1}^n(x_i-\overline{X}{)}^2$
• D. $\frac{1}{n}{\sum }_{i=1}^n|x_i-\overline{X}|$

Answer 1

The correct option is D

$\frac{1}{n}{\sum }_{i=1}^n|x_i-\overline{X}|$

The mean deviation for n observations

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