Question

Given n observations x₁, x₂ ......x_n and their central tendency a. Find the mean deviation about a.

• A. M.D(a) = $\frac{\sum _{i=1}^n|x_i-a|}{n}$
• B. M.D(a) = $\frac{\sum _{i=1}^n{x}_i-a}{n}$
• C. M.D(a) = $\sum _{i=1}^n{x}_i-a$
• D. a and b

Answer 1

The correct option is A

M.D(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) = $\frac{\sum _{i=1}^n|x_i-a|}{n}$