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Find the standard deviation of first n natural numbers. |
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Answer» `Sigmax_(i)^(2)=1^(2)+2^(2)+3^(3)+…..+n^(2)=(n(n+1)(2N+1))/(6)` `sigma=sqrt((Sigmax_(i)^(2))/N-(Sigma_(x_(i))/N)^(2)` `sqrt((n(n+1)(2n+1))/(6N)-(n^(2)(n+1)^(2))/(4N^(2))` `sqrt(((n+1)(2n+1))/(6)-((n+1)^(2))/(4)` `sqrt((2(2n^(2)+3n+1)-3(n^(2)+2n+1))/(12)` `sqrt((4n^(2)+6n+2-3n^(2)-6n-3)/(12)` `sqrt((n^(2)-1)/12)` |
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