Question

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

Answer 1

The mean of first n natural numbers is calculated as follows.

\(Mean=\frac{sum\,of\,all\,observations}{Number\,of\,all\,observations}\)

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

Variance(σ²)= \(\frac{1}{2}\sum_{i=1}^n(x_i-\bar{x})^2\)

=\(\frac{1}{2}\sum_{i=1}^nx_i^2-\frac{1}{n}\sum_{i=1}^n2(\frac{n+1}{2})X_i+\frac{1}{n}\sum_{i=1}^{n}(\frac{n+1}{2})^2\)

=\(\frac{1}{2}\frac{n(n+1)(2n+1)}{6}-(\frac{n+1}{n})[\frac{n(n+1)}{2}]+\frac{(n+1)^2}{4n}×n\)

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

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

\(=(n+1)[\frac{4n+2-3n-3}{12}]\)

\(\frac{(n+1)(n-1)}{12}\)

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