Question

Assertion :The variance of first $n$ natural numbers is $\frac{n^2-1}{6}$ Reason: The sum and the sum of squares of first $n$ natural numbers are $\frac{n(n+1)}{2}$ and $\frac{n(n+1)(2n+1)}{6}$ respectively.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

First $n$ natural numbers are, $1,2,3,..........n$
Here $\sum x_i=\frac{n(n+1)}{2},\sum x_i^2=\frac{n(n+1)(2n+1)}{6}$
Ergo variance is ${\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2=\frac{n^2-1}{12}$
Clearly Assertion is incorrect, obviously Reason is correct.