Question

The variance of first n natural numbers is

• A.
• B.
• C.
• D. none of these

Answer 1

The correct option is B

$Wehave{\sigma }^2=\frac{1}{n}{\sum }_i^n{=}_1{x}_i^2-{\left[\frac{1}{n}{\sum }_i^n{=}_1{x}_i\right]}^2=\frac{1}{n}(1^2+2^2+......+n^2)-{\left[\frac{1}{n}\right(1+2+....+n\left)\right]}^2=\frac{1}{n}\frac{n(n+1)(2n+1)}{6}-{\left[\frac{n+1}{2}\right]}^2=\frac{n^2-1}{12}$