Question

If $\overline{X}$ is the mean of $x_1,x_2,x_3,...,x_n$. Then, the algebraic sum of the deviations about mean $\overline{X}$ is

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

Answer 1

Answer: (A)

$0$

Mean = $\overline{X}=\frac{x_1+x_2+x_3+.....+x_n}{n}$
We have been asked for the algebraic sum of the deviations of each of the observations from the mean, i.e. $(x_1-\overline{X})+(x_2-\overline{X})+(x_3-\overline{X})+....+(x_{n-1}-\overline{X})+(x_n-\overline{X})=x_1+x_2+x_3+....+x_n-n\overline{X}$
$=n\overline{X}-n\overline{X}$
$=0$
Hence, option 'A' is correct.