Find the mean and variance for the first n na | vedclass

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Question

(A) The mean of the first $n$ natural numbers is calculated as follows:Mean $= \frac{	ext{Sum of all observations}}{	ext{Number of observations}} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The mean of the first $n$ natural numbers is calculated as follows:Mean $= \frac{	ext{Sum of all observations}}{	ext{Number of observations}} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$Variance $(\sigma^2)$ is given by $\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$.Using the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$:$\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$$\sigma^2 = \frac{1}{n} \left[ \frac{n(n+1)(2n+1)}{6} ight] - \left( \frac{n+1}{2} ight)^2$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$$\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$$\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12}$$\sigma^2 = \frac{(n+1) [4n+2 - 3n-3]}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$Thus,the mean is $\frac{n+1}{2}$ and the variance is $\frac{n^2-1}{12}$.

Find the mean and variance for the first $n$ natural numbers.

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