Question

Given n observations x₁, x₂ ......x_n and their central tendency a, the mean deviation about a, M.D(a) =

• A. $\frac{{\sum }_{i=1}^n|x_i-a|}{n}$
• B. $\frac{{\sum }_{i=1}^n{x}_i-a}{n}$
• C. ${\sum }_{i=1}^n{x}_i-a$

Answer 1

The correct option is A $\frac{{\sum }_{i=1}^n|x_i-a|}{n}$

Absolute deviations of observations about a will be

$|$ $x_1$ - a $|$, $|$ $x_2$ - a $|$, $|$ $x_3$ - a $|$....,$|$ $x_n$ - a $|$

$\therefore$ Mean of absolute deviations = $\frac{|x_1-a|+|x_2-a|+....+|x_n-a|}{n}$

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

$\therefore$ MD(a) = $\frac{{\sum }_{i=1}^n|x_i-a|}{n}$