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MHTCET 15 May 2024 Shift 2 Paper

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Question : 135 of 150

Marks: +1, -0

Variance of first n natural numbers is

\_\_\_\_

.

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

Solution:

Concept:

For

n

numbers, variance is

\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2

.

Explanation:

For first

n

natural numbers,

\sum x_i = 1+2+\dots+n = \frac{n(n+1)}{2}

.

Also,

\sum x_i^2 = 1^2+2^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6}

.

Substitute into the variance formula:

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

Simplifying:

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

\sigma^2 = \frac{n+1}{2}\left(\frac{2n+1}{3} - \frac{n+1}{2}\right)

\sigma^2 = \frac{n+1}{2} \cdot \frac{n-1}{6} = \frac{n^2-1}{12}

Answer:

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

, i.e. Option D.

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