Calculate the mean and standard deviation of first natural numbers.

Options

A. `sigma=(N^(2)-1)/(12)`
B. `sigma=sqrt((N^(2)-1)/(12))`
C. `sigma=sqrt((N-1)/(12))`
D. `sigma=sqrt((N^(2)-1)/(6N))`

Answer

B

Explanation

`sigma^(2)=(1)/(n)sum_(i=1)^(n)x_(i)^(2)-((1)/(n)sum_(i=1)^(n)x_(i))^(2)`
`=(1)/(2)(1^(2)+2^(2)+3^(2)+.....+n^(2))-((1)/(n)(1+2+3+.....+n))^(2)`
`=(1)/(n)(n(n+1)(2n+1))/(6)-((1)/(n).(n(n+1))/(2))^(2)`
`=(n^(2)-1)/(12)" "therefore sigma=sqrt((n^(2)-1)/(12)).`