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, $\sigma =\sqrt{\frac{\Sigma x_i^2}{N}-(\bar{X}{)}^2}$
$\therefore \bar{X}=\frac{\Sigma X_i}{N}$
$=\frac{1+3+5+\dots (2n-1)}{n}$
$=\frac{\frac{n}{2}[1+2n-1]}{n}$
$=\frac{n^2}{n}=n$
Again, $\Sigma x_i^2=1^2+3^2+5^2+\dots (2n-1{)}^2$
$=\Sigma (2n-1{)}^2$
$=\Sigma \left(4n^2-4n+1\right)$
$=4\Sigma n^2-4\Sigma n+\Sigma 1$
$=\frac{4n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}+n$
$=n\left[\frac{2}{3}(n+1)2n+1\right)-2(n+1)+1]$
$=\frac{n}{3}\left[2\left(2n^2+3n+1\right)-6(n+1)+3\right]$
$=\frac{n}{3}\left[4n^2+6n+2-6n-6+3\right]$
$=\frac{n}{3}\left[4n^2-1\right]$
$\therefore \sigma =\sqrt{\frac{n\left(4n^2-1\right)}{3n}-n^2}$
$=\sqrt{\frac{4n^2-1}{3}-n^2}$
$=\sqrt{\frac{4n^2-1-3n^2}{3}}$
$=\sqrt{\frac{n^2-1}{3}}$