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Q.Find the variance of first n natural numbers.

West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 2mImportance★★★★★est
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Using Var=∑xi2n−(∑xin)2\text{Var}=\dfrac{\sum x_i^2}{n}-\left(\dfrac{\sum x_i}{n}\right)^2 with the standard sums for 1,2,…,n1,2,\dots,n gives n2−112\dfrac{n^2-1}{12}.

For the data 1,2,…,n1,2,\dots,n:

Mean=xˉ=∑i=1nin=n(n+1)/2n=n+12.\text{Mean}=\bar x=\dfrac{\sum_{i=1}^n i}{n}=\dfrac{n(n+1)/2}{n}=\dfrac{n+1}{2}.

∑i=1ni2n=n(n+1)(2n+1)/6n=(n+1)(2n+1)6.\dfrac{\sum_{i=1}^n i^2}{n}=\dfrac{n(n+1)(2n+1)/6}{n}=\dfrac{(n+1)(2n+1)}{6}.

Variance =∑xi2n−xˉ2=(n+1)(2n+1)6−(n+12)2=\dfrac{\sum x_i^2}{n}-\bar x^2=\dfrac{(n+1)(2n+1)}{6}-\left(\dfrac{n+1}{2}\right)^2 …

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