Question

If the mean of a observations $x_1,x_2,.....x_{n}is\overline{x},$ then the sum of deviations of observations from mean is

• A. 0
• B. $n\overline{x}$
• C. $\frac{\overline{x}}{n}$
• D. None of these

Answer 1

The correct option is A 0

Given $x_1,x_2,x_3,.....x_n$ are observations
$\frac{x_1+x_2+....+x_n}{n}=\overline{x}$
$\Rightarrow x_1+x_2+...+x_n=n\overline{x}$ .....(1)
Now deviations of $x_1,x_2,...x_n$ from mean is $x_1-\overline{x},x_2-\overline{x},.....x_n-\overline{x}$
$\therefore$ req.sum $=(x_1-\overline{x})+\left(x_2\overline{x}\right)+....+(x_n-\overline{x})=(x_1+x_2+...+x_n)-n\overline{x}$
$=n\overline{x}-n\overline{x}=0$ (by (1))