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Question

Mathematics Question on Variance and Standard Deviation

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

Answer

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

Mean=sum of all observationsNumber of all observationsMean=\frac{sum\,of\,all\,observations}{Number\,of\,all\,observations}

Mean = n(n+1)n=n+12\frac{n(n+1)}{n}=\frac{n+1}{2}

Variance(σ2)= 12∑i=1n(xi−xˉ)2\frac{1}{2}\sum_{i=1}^n(x_i-\bar{x})^2

=12∑i=1nxi2−1n∑i=1n2(n+12)Xi+1n∑i=1n(n+12)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

=12n(n+1)(2n+1)6−(n+1n)[n(n+1)2]+(n+1)24n×n\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

==(n+1)(2n+1)6−(n+1)24+(n+1)24=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}+\frac{(n+1)^2}{4}

=(n+1)(2n+1)6−(n+1)24=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}

=(n+1)[4n+2−3n−312]=(n+1)[\frac{4n+2-3n-3}{12}]

(n+1)(n−1)12\frac{(n+1)(n-1)}{12}

=n2−112=\frac{n^2-1}{12}