Question

The standard deviation of the first n natural numbers is

• A. $\frac{n+1}{2}$
  
• B. $\sqrt{\frac{n(n+1)}{2}}$
  
• C. $\sqrt{\frac{n^2-1}{12}}$
  
• D. none of these

Answer 1

The correct option is C

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

S.D. of the first n natural numbers
$\sqrt{\frac{1}{n}\sum x^2-{\left(\frac{\sum x}{n}\right)}^2}[\because \overline{x}=\frac{\sum x}{n}]=\sqrt{\frac{n(n+1)(2n+1)}{6n}-{\left(\frac{n(n+1)}{2n}\right)}^2}=\sqrt{\frac{(n+1)(2n+1)}{6}-{\left(\frac{(n+1)}{2}\right)}^2}=\sqrt{\frac{n+1}{2}(\frac{2n+1}{3}-\frac{n+1}{2})}=\sqrt{\frac{n+1}{2}\left(\frac{4n+2-3n-3}{6}\right)}=\sqrt{\frac{n^2-1}{12}}$