Question

The standard deviation of the first n natural numbers is

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

Answer 1

Answer: (C)

Since, $\Sigma n=\frac{n(n+1)}{2}$ and $\Sigma n^2=\frac{n(n+1)(2n+1)}{6}$
$\therefore$ $SD=\sqrt{\frac{\Sigma x^2}{n}-{\left(\Sigma \frac{x}{n}\right)}^2}$
$=\sqrt{\frac{\Sigma n^2}{n}-{\left(\frac{\Sigma n}{n}\right)}^2}$
$=\sqrt{\frac{n(n+1)(2n+1)}{6n}-{\left(\frac{n+(n+1)}{2n}\right)}^2}$
$=\sqrt{\frac{(n+1)(2n+1)}{6}-\frac{{(n+1)}^2}{4}}$
$=\sqrt{\frac{n^2-1}{12}}$