Question

variance of first n odd natural numbers and



variance of first n even numbers

Answer 1

odd variance=-(n2+1)/3
even variance=(n2-1)/3

Answer 2

The variance of first n odd natural numbers is $\frac{n^{2}-1 }{3}$ .

The variance of first n even natural numbers is $\frac{n^{2}-1 }{3}$ .

Step-by-step explanation:

We have to calculate the variance of first n odd natural numbers;

As we know that the formula for variance is given by;

    Variance, $\sigma^{2}$  =  $\frac{\sum X_i^{2} }{n}- \bar X^{2}$

The series of first n odd natural numbers is;

1, 3, 5, 7,........., (2n-1)

So, firstly mean of this series is calculated as;

    $\bar X$  =  $\frac{1 + 3+5+......+(2n-1)}{n}$

As we can see that the numerator is a series of an AP, so the sum of n terms of an AP is given as  $S_n=\frac{n}{2}[2a+(n-1)d]$

Here, a = 1 and d = 2

So,   $\bar X = \frac{\frac{n}{2}[2\times 1+(n-1)\times 2] }{n}$

        $\bar X = \frac{\frac{n}{2}[2 + 2n-2] }{n}$

        $\bar X = \frac{n^{2} }{n}$ = n

Now,  $\sum X_i^{2} = 1^{2} +2^{2} +3^{2} +........+(2n-1)^{2}$

We know that sum of squares of odd natural numbers is given by ;

       $\sum X_i^{2}=\frac{n(2n+1)(2n-1)}{3}$

Now, putting these values in variance formula we get;

               Variance, $\sigma^{2}$  =  $\frac{n(2n+1)(2n-1)}{3n}-n^{2}$

                                     =  $\frac{4n^{2}-2n+2n-1 }{3}-n^{2}$

                                     =  $\frac{4n^{2}-3n^{2} -1 }{3}$

                                     =  $\frac{n^{2}-1 }{3}$

So, the variance of first n odd natural numbers is $\frac{n^{2}-1 }{3}$ .

• Now, we will calculate the variance of first n even numbers;

The series of first n even natural numbers is;

2, 4, 6, 8,.........,2n

So, firstly mean of this series is calculated as;

    $\bar X$  =  $\frac{2 +4+6+......+2n}{n}$

As we can see that the numerator is a series of an AP, so the sum of n terms of an AP is given as  $S_n=\frac{n}{2}[2a+(n-1)d]$

Here, a = 2 and d = 2

So,   $\bar X = \frac{\frac{n}{2}[2\times 2+(n-1)\times 2] }{n}$

        $\bar X = \frac{\frac{n}{2}[4 + 2n-2] }{n}$

        $\bar X = \frac{n(n+1) }{n}$ = n + 1

Now,  $\sum X_i^{2} = 2^{2} +4^{2} +6^{2} +........+(2n)^{2}$

Taking $2^{2}$ common we get;

$\sum X_i^{2} = 2^{2}[ 1^{2} +2^{2} +3^{2} +........+(n)^{2}]$

We know that sum of squares of n natural numbers is given by ;

                    =  $\frac{n(n+1)(2n+1)}{6}$

So,  $\sum X_i^{2} = \frac{4n(n+1)(2n+1)}{6}$

Now, putting these values in variance formula we get;

               Variance, $\sigma^{2}$  =  $\frac{4n(n+1)(2n+1)}{6n}-(n+1)^{2}$

                                     =  $\frac{2(n+1)(2n+1)}{3}-(n+1)^{2}$

                                     =  $\frac{n+1}{3} [2(2n+1)-3(n+1)]$

                                     =  $\frac{n+1}{3} [4n+2-3n-3]$

                                     =   $\frac{n+1}{3} (n-1)$

                                     =  $\frac{n^{2}-1 }{3}$

So, the variance of first n even natural numbers is $\frac{n^{2}-1 }{3}$ .