find mean variance standard deviation of first n natural numbers
Answers
Answered by
3
First n natural number 1,2,3,4,5,6......n.
Mean=1+2+3+4+5+6+.........n.
=n(n+1)/2/n
Variance=sigma(x1-x2_(bar))^2
As the calculation is very large, so
Variance=sigma(xi)^2/n-Mean
As,1^2+2^2+3^2+4^2....n^2=n(n+1)(2n+1)/6
Therefore Variance=n(n+1)(2n+1)/6n-(n+1)^2/4
= (n+1)(2n+1)/6-(n+1)^2/4
=(n+1)/2(2n+1/3-n+1/2)
=(n+1)/2(4n+2-3n-3/6)
=(n+1)/2(n-1/6)
=n^2-1/12 is the ans.
Hope This HELPS☺️☺️
Mean=1+2+3+4+5+6+.........n.
=n(n+1)/2/n
Variance=sigma(x1-x2_(bar))^2
As the calculation is very large, so
Variance=sigma(xi)^2/n-Mean
As,1^2+2^2+3^2+4^2....n^2=n(n+1)(2n+1)/6
Therefore Variance=n(n+1)(2n+1)/6n-(n+1)^2/4
= (n+1)(2n+1)/6-(n+1)^2/4
=(n+1)/2(2n+1/3-n+1/2)
=(n+1)/2(4n+2-3n-3/6)
=(n+1)/2(n-1/6)
=n^2-1/12 is the ans.
Hope This HELPS☺️☺️
Similar questions