Question

Standard deviation of first $n$ odd natural numbers is

• A. $\sqrt{n}$
• B. $\sqrt{ \frac{(n+2)(n+1)}{3}}$
• C. $\sqrt{ \frac{n^2 - 1 }{3}}$
• D. $n$
• E. $2n$

Answer 1

Answer: (C)

Standard deviation, $\sigma=\sqrt{\frac{\Sigma x_{i}^{2}}{N}-(\bar{X})^{2}}$
$\therefore \bar{X}=\frac{\Sigma X_{i}}{N}$
$=\frac{1+3+5+\ldots(2 n-1)}{n}$
$=\frac{\frac{n}{2}[1+2 n-1]}{n}$
$=\frac{n^{2}}{n}=n$
Again, $\Sigma x_{i}^{2}=1^{2}+3^{2}+5^{2}+\ldots(2 n-1)^{2}$
$=\Sigma(2 n-1)^{2}$
$=\Sigma\left(4 n^{2}-4 n+1\right)$
$=4 \Sigma n^{2}-4 \Sigma n+\Sigma 1$
$=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$
$\left.=n\left[\frac{2}{3}(n+1) 2 n+1\right)-2(n+1)+1\right]$
$=\frac{n}{3}\left[2\left(2 n^{2}+3 n+1\right)-6(n+1)+3\right]$
$=\frac{n}{3}\left[4 n^{2}+6 n+2-6 n-6+3\right]$
$=\frac{n}{3}\left[4 n^{2}-1\right]$
$\therefore \sigma=\sqrt{\frac{n\left(4 n^{2}-1\right)}{3 n}-n^{2}}$
$=\sqrt{\frac{4 n^{2}-1}{3}-n^{2}}$
$=\sqrt{\frac{4 n^{2}-1-3 n^{2}}{3}}$
$=\sqrt{\frac{n^{2}-1}{3}}$