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 $=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 \cdot n}$
$ \Rightarrow \,\frac{n+1}{2}$
So, coefficient of variance $=\frac{\sigma_{X}}{\bar{x}} \times 100$
$\sigma_{x}=\sqrt{\frac{\Sigma n^{2}}{n}-\left(\frac{\Sigma n}{n}\right)^{2}}$
$\Rightarrow \, \sigma_{x}=\sqrt{\frac{n(n+1)(2 n+1)}{6 \cdot n}-\left[\frac{n(n+1)}{2 \cdot n}\right]^{2}}$
$\Rightarrow \, \sigma_{x}=\sqrt{\frac{n+1}{2}\left\{\frac{2 n+1}{3}-\frac{n+1}{2}\right\}}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left[\frac{4 n+2-3 n-3}{6}\right]}$
$=\sqrt{\left(\frac{n+1}{2}\right)\left(\frac{n-1}{6}\right)}=\sqrt{\frac{n^{2}-1}{12}}$
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}}$