Q. If $\overline{x}$ is the arithmetic mean of $n$ independent variates $x_{1},x_{2},x_{3},\ldots,x_{n}$ each of the standard derivation $\sigma$, then variance $(\overline{x})$ is
A) $\frac{\sigma^{2}}{n}$
B) $\frac{{n\sigma}^{2}}{2}$
C) $\frac{(n + 1)\sigma^{2}}{3}$
D) None of these
Explanation
(a)
We have, $\overline{x} = \frac{1}{n}\sum_{i = 1}^{n}x_{i}$
$\therefore\ \ \ var\left( \overline{x} \right) = \frac{1}{n^{2}}\left\lbrack \sum_{i = 1}^{n}{var(x_{i})} + 2\sum_{i \neq j}^{n}{cov(x_{i},\ x_{j})} \right\rbrack$
$= \frac{1}{n^{2}}\left\lbrack n\sigma^{2} \right\rbrack = \frac{\sigma^{2}}{n}$
[$\because x_{i}$ and $x_{j}$ are independent variable, therefore $cov\left( x_{i},\ x_{j} \right) = 0$]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.