Question

The variance of the first n natural numbers is

• A. $\frac{n^{2} - 1}{12}$
• B. $\frac{n^{2} - 1}{6}$
• C. $\frac{n^{2} + 1}{6}$
• D. $\frac{n^{2} + 1}{12}$

Answer 1

Answer: (A)

$\frac{n^{2} - 1}{12}$

Answer 2

Variance $= (\text{S.D.})^{2} = \frac{1}{n}\Sigma x^{2} - \left( \frac{\Sigma x}{n} \right)^{2}$, $\left( \because\mspace{6mu}\mspace{6mu}\bar{x} = \frac{\Sigma x}{n} \right)$

$= \frac{n(n + 1)\mspace{6mu}(2n + 1)}{6n} - \left( \frac{n(n + 1)}{2n} \right)^{2} = \frac{n^{2} - 1}{12}$.