Question

The sum of the deviations of the variates from the arithmetic mean is always

• A. $1$
• B. $0$
• C. $-1$
• D. Real number

Answer 1

The correct option is B

$0$

Explanation for the correct answer:

Let $\overline{x}$ be the arithmetic mean of $x_1,x_2,x_3,......,x_n$

$(x_i-\overline{x})$ is the deviation of the $i^{th}$ term from the mean.

We take a summation of all these deviations

$\sum _{i=1}^n(x_i-\overline{x})=\sum _{i=1}^n{x}_i-\underset{i=1}{\overset{n}{\sum \overline{x}}}$

$=\sum _{i=1}^n{x}_i-n\overline{x}$

$=\sum _{i=1}^n{x}_i-n\left(\frac{1}{n}\underset{i=1}{\overset{n}{\sum x_i}}\right)$ …[by definition of arithmetic mean $\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathit{x}}_{\mathbf{i}}$]

$=\sum _{i=1}^n{x}_i-\underset{i=1}{\overset{n}{\sum x_i}}$

$\Rightarrow$ $\sum _{i=1}^n(x_i-\overline{x})=0$

Hence, the sum of all deviations of all the variates from the arithmetic mean is $0$.

Hence, option B is the correct answer.