# Find the standard deviation of the first n natural numbers.

$\begin{array}{1 1}(A)\;\sqrt {\large\frac{n^2-1}{12}}\\(B)\;(n+1)\\(C)\;0\\(D)\;\large\frac{n-1}{12}\end{array}$

Toolbox:
• The formula to calculate SD is $SD= \sqrt { \large\frac{\sum x^2}{n} +\bigg(\large\frac{\sum x}{n} \bigg)^2}$
The first n natural numbers are $1,2,3........n$
As we know sum of n natural numbers are $\sum x =\large\frac{n(n+1)}{2}$
$\sum x^2 = \sum (1^2+2^2+3^2+.........+n^2)$
Sum of square numbers $\sum x^2 =\large\frac{n(n+1)(2n+1)}{6}$
$SD= \sqrt {\large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n}\bigg)^2}$
$\quad= \sqrt { \large\frac{n(n+1)(2n+1)}{6n} - \bigg( \large\frac{n(n+1)}{2n} \bigg)^2 }$
$\quad =\sqrt { \large\frac{(n+1)(2n+1) }{6} - \bigg( \large\frac{(n+1)}{2}\bigg)^2}$
$\qquad= \sqrt { (n+1) \bigg[ \large\frac{2n+1}{6} -\frac{(n+1)}{4} \bigg]}$
$\qquad= \sqrt {\large\frac{(n+1) (n-1)}{12}}$
$\qquad= \sqrt {\large\frac{n^2 -1}{12}}$
Hence A is the correct answer.
i think the formula mentioned in the tool box is wrong.Please modify and
SD=root ( ∑x^2/n+∑x/n )