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UPSC NDA Math Model Paper 7

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Question : 33 of 120

Marks: +1, -0

The variance first n natural numbers is:

$\frac{n^{2}+1}{12}$
$\frac{n^{2}-1}{12}$
$\frac{(n+1)^{2}}{12}$
$\frac{2n^{2}-1}{8}$

Solution:

To Find: Variance first n natural numbers

\operatorname{Var}(X)=E(X^{2})-(E(X))^{2}

Now

\operatorname{E}(X^{2}\text{right})=

Mean of square of 'n' natural numbers

=\frac{\text{Sum of square of first n natural numbers}}{n}

=\frac{n(n+1)(2n+1)}{6n}

=\frac{(n+1)(2n+1)}{6}

\operatorname{E}(X)=

Mean of 'n' natural numbers

=\frac{\text{sum of all terms}}{\text{number of terms}}

=\frac{n(n+1)}{2n}=\frac{(n+1)}{2}

(\operatorname{E}(X))^{2}=\left[\frac{(n+1)}{2}\right]^{2}

\operatorname{Var}(X)=\operatorname{E}(X^{2})-(\operatorname{E}(X))^{2}

=\frac{(n+1)(2n+1)}{6}-\left(\frac{(n+1)}{2}\right)^{2}

=\frac{(n+1)}{2}\left[\frac{(2n+1)}{3}-\frac{(n+1)}{2}\right]

=\frac{(n+1)}{2}\left[\frac{4n+2-3n-3}{6}\right]

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

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