Question

The standard deviation $\sigma$ of the first $N$ natural numbers can be obtained using which one of the following formula?

• A. $\sigma =\frac{N^2-1}{12}$
• B. $\sigma =\sqrt{\frac{N^2-1}{12}}$
• C. $\sigma =\sqrt{\frac{N-1}{12}}$
• D. $\sigma =\sqrt{\frac{N^2-1}{6N}}$

Answer 1

The correct option is C $\sigma =\sqrt{\frac{N-1}{12}}$

${\sigma }^2=\frac{1}{n}{\sum }_{i=1}^n{i}^2-{\left(\frac{1}{n}{\sum }_{i=1}^{n}i\right)}^2$

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

$=\frac{1}{n}\times \frac{n(n+1)(2n+1)}{6}-(\frac{n+1}{2}{)}^2=\frac{n^2-1}{12}$

$\therefore \sigma =\sqrt{\frac{n^2-1}{12}}$