InterviewSolution
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InterviewSolution
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Find the variance of first n natural numbers. |
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Answer» Here, the variable are1,2,3,…, n ∴ Mean, \(\bar{x}=\frac{\sum x_i}{n}\) \(=\frac{1+2+3+...+n}{n}\) \(=\frac{n(n+1)}{2n}\) \(=\frac{n+1}{2}\) Variance = \(\sum x_i^2-(\bar{x})^2\) = \(\frac{1^2+2^2+3^2+...+n^2}{n}-(\frac{n+1}{2})^2\) = \(\frac{n(n+1)(2n+1)}{6n}-(\frac{n+1}{2})\) = \((n+1)\{\frac{4n+2-3n-3}{12}\}\) = \(\frac{(n+1)(n-1)}{12}\) = \(\frac{n^2-1}{12}\) |
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