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  • Find the mean and variance for each of the data. (2) First n natural numbers

2.   Find the mean and variance for each of the data.

    First n natural numbers.
 

Answers (1)

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Mean (\overline{x}) of first n natural numbers:

\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}

We know, Variance\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2 

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

We know that (a-b)^2 = a^2 - 2ab + b^2

\\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}

\\ \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}

\\ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}

Hence, Mean = \frac{n+1}{2} and Variance = \frac{n^2-1}{12}

Posted by

HARSH KANKARIA

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