Question

The coefficient of variation of the first $n$ natural numbers is

• A. $\frac{100}{\sqrt{3}}(n-1)$
• B. $\frac{100}{\sqrt{3}}\sqrt{\frac{n+1}{n-1}}$
• C. $\frac{\sqrt{3}}{100}\sqrt{\frac{n+1}{n-1}}$
• D. $\frac{100}{\sqrt{3}}\sqrt{\frac{n-1}{n+1}}$

Answer 1

Answer: (D)

The coefficient of variation of first $n$ natural number is
Mean $=\frac{1+2+3+\dots +n}{n}=\frac{n(n+1)}{2\cdot n}$
$\Rightarrow \frac{n+1}{2}$
So, coefficient of variance $=\frac{{\sigma }_X}{\bar{x}}\times 100$
${\sigma }_x=\sqrt{\frac{\Sigma n^2}{n}-{\left(\frac{\Sigma n}{n}\right)}^2}$
$\Rightarrow {\sigma }_x=\sqrt{\frac{n(n+1)(2n+1)}{6\cdot n}-{\left[\frac{n(n+1)}{2\cdot n}\right]}^2}$
$\Rightarrow {\sigma }_x=\sqrt{\frac{n+1}{2}\left\{\frac{2n+1}{3}-\frac{n+1}{2}\right\}}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{4n+2-3n-3}{6}\right]}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left(\frac{n-1}{6}\right)}=\sqrt{\frac{n^2-1}{12}}$
So, coefficient of variance
$=\frac{\sqrt{\frac{n^2-1}{12}}}{\frac{n+1}{2}}\times 100\sqrt{\frac{n^2-1}{(n+1{)}^2}}\cdot \frac{100}{\sqrt{3}}$
$\Rightarrow \frac{100}{\sqrt{3}}\sqrt{\frac{n-1}{n+1}}$