The coefficient of variation of the first n n | vedclass

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(D) The mean of the first $n$ natural numbers is $\bar{x} = \frac{n+1}{2}$.The variance $\sigma^2$ of the first $n$ natural numbers is given by $\sigm

Vedclass · Accepted Answer

$\frac{100}{\sqrt{3}} \sqrt{\frac{n-1}{n+1}}$

Vedclass · Answer

(D) The mean of the first $n$ natural numbers is $\bar{x} = \frac{n+1}{2}$.The variance $\sigma^2$ of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2-1}{12}$.Thus,the standard deviation is $\sigma = \sqrt{\frac{n^2-1}{12}}$.The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} 	imes 100$.Substituting the values:$CV = \frac{\sqrt{\frac{n^2-1}{12}}}{\frac{n+1}{2}} 	imes 100 = \frac{\sqrt{\frac{(n-1)(n+1)}{12}}}{\frac{n+1}{2}} 	imes 100$$= \sqrt{\frac{(n-1)(n+1)}{12} 	imes \frac{4}{(n+1)^2}} 	imes 100 = \sqrt{\frac{n-1}{3(n+1)}} 	imes 100$$= \frac{100}{\sqrt{3}} \sqrt{\frac{n-1}{n+1}}$.

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

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