Question

Variance of the first $11$ natural numbers is:
A. $\sqrt{5}$
B. $\sqrt{10}$
C. $5\sqrt{2}$
D. $10$

Answer 1

Hint: Here we have to find the variance of first $11$ natural numbers. We will firstly find the variance when the number of observations is $n$ by using the mean of the natural number formula in the variance formula and also the sum of the $n$ numbers formula for simplifying it further. Finally after simplification we will put $n=11$ and simplify it to get the desired answer.

Complete step by step answer:
We have to find the variance of the first $11$ numbers.
The formula for calculating variance is as follows:
$\sigma =\dfrac{\sum{{{\left( {{x}_{i}} \right)}^{2}}}}{n}-{{\mu }^{2}}$….$\left( 1 \right)$
Where, \[\sigma =\] Variance and $\mu =$ Mean
We know that the mean of natural number is,
Mean $=\dfrac{n+1}{2}$
Also we can write the value ${{\sum{\left( {{x}_{i}} \right)}}^{2}}$ as follows for $n$ numbers,
${{\sum{\left( {{x}_{i}} \right)}}^{2}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.......{{n}^{2}}$
On substituting the above two values in formula (1) we get,
$\Rightarrow \sigma =\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.......{{n}^{2}}}{n}-{{\left( \dfrac{n+1}{2} \right)}^{2}}$
Now as we know sum of square of $n$ numbers is equal to $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ put it above,
$\Rightarrow \sigma =\dfrac{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}{n}-{{\left( \dfrac{n+1}{2} \right)}^{2}}$
$\Rightarrow \sigma =\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6n}-{{\left( \dfrac{n+1}{2} \right)}^{2}}$
Simplifying it further we get,
$\Rightarrow \sigma =\dfrac{n\times 2n+n\times 1+1\times 2n+1\times 1}{6}-\dfrac{{{n}^{2}}+1+2n}{4}$
$\Rightarrow \sigma =\dfrac{2{{n}^{2}}+n+2n+1}{6}-\dfrac{{{n}^{2}}+1+2n}{4}$
Taking L.C.M both sides we get,
$\Rightarrow \sigma =\dfrac{2\left( 2{{n}^{2}}+3n+1 \right)-3\left( {{n}^{2}}+1+2n \right)}{12}$
$\Rightarrow \sigma =\dfrac{4{{n}^{2}}+6n+2-3{{n}^{2}}-3-6n}{12}$
Simplify it further,
$\Rightarrow \sigma =\dfrac{{{n}^{2}}-1}{12}$
Put $n=11$ ,
$\Rightarrow \sigma =\dfrac{{{11}^{2}}-1}{12}$
$\Rightarrow \sigma =\dfrac{121-1}{12}$
So we get,
$\Rightarrow \sigma =\dfrac{120}{12}$
$\Rightarrow \sigma =10$
So we get the variance of the first $11$ natural number as $10$ .

So, the correct answer is “Option D”.

Note:
Variance denotes how the data is spread out from the mean or how data points differ from the mean. It depends on the standard deviation as it is equal to the square of standard deviation. In this question as we have to find the variance of the first $11$ natural number we can take those numbers and put it in the formula directly but that can have a chance of miscalculation as there will be $11$ numbers whose square of the difference with the mean is to be calculated. So getting a general formula is a better approach for this question.