Q. The variance of the first $n$ natural numbers is
A) $\left( \frac{n^{2} - 1}{12} \right)$
B) $\frac{n(n^{2} - 1)}{12}$
C) $\left( \frac{n^{2} + 1}{12} \right)$
D) $\frac{n(n^{2} + 1)}{12}$
Explanation
(a)
$\overline{x} = \frac{1 + 2 + 3 + ... + n}{n} = \frac{(n + 1)}{2}$
Variance, $\sigma^{2} = \frac{\sum_{}^{}\left( x_{i} \right)^{2}}{n} - \left( \overline{x} \right)^{2}$
$= \frac{\sum_{}^{}n^{2}}{n} - \left( \frac{n + 1}{2} \right)^{2}$
$= \frac{n(n + 1)(2n + 1)}{6n} - \left( \frac{n + 1}{2} \right)^{2}$
$= \frac{n^{2} - 1}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.