Question

The variance of first $n$ natural numbers, is
A. $\dfrac{{n + 1}}{2}$
B. $\dfrac{{{n^2} + 1}}{{12}}$
C. $\dfrac{{{n^2} - 1}}{6}$
D. $\dfrac{{{n^2} - 1}}{{12}}$

Answer 1

Hint: For required solution, we will see first $n$ natural numbers and then we have to calculate the variance of those numbers. Now, to calculate variance we need to find mean of all the $n$ natural numbers. Now, we know, variance can be calculated using the formula ${\sigma ^2} = \dfrac{1}{n}\sum {x_i^2 - {{\left( {\bar x} \right)}^2}} $ , where $\bar x$ is the mean of the $n$ natural numbers and $n$ is the total count of the numbers.

Complete step-by-step answer:
Now, we know,
The first $n$ natural numbers are = 1, 2, 3, 4, 5….. $n$
The count of these numbers is $n$
Now, to find mean we have to use the formula,
 Mean, $\bar x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}$
So, putting values in the formula,
\[\bar x = \dfrac{{1 + 2 + 3 + 4 + 5....n}}{n}\]
Now, using formula to find sum of \[n\] natural numbers
$s = \dfrac{{n\left( {n + 1} \right)}}{2}$
Now, putting the above value in mean,
\[
   \Rightarrow \bar x = \dfrac{{n\left( {n + 1} \right)}}{{2 \times n}} \\
   \Rightarrow \bar x = \dfrac{{\left( {n + 1} \right)}}{2} \\
 \]
Now, ${\bar x^2} = \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}.............\left( 1 \right)$
Now, we know that the sum of the squares of \[n\] natural numbers is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
So, $\dfrac{1}{n}\sum {x_i^2 = } \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{n \times 6}}.................\left( 2 \right)$

Now, putting values of equation (1) and (2) in variance formula,
\[ \Rightarrow {\sigma ^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}\]
Now, on simplifying the equation,
\[
   \Rightarrow {\sigma ^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{4} \\
   \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right) \\
   \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{4n + 2 - 3n - 3}}{6}} \right) \\
   \Rightarrow {\sigma ^2} = \dfrac{{\left( {n + 1} \right)}}{2}\left( {\dfrac{{n - 1}}{6}} \right) \\
   \Rightarrow {\sigma ^2} = \dfrac{{{n^2} - 1}}{{12}} \\
 \]
Hence, the variance of first $n$ natural numbers is \[\dfrac{{{n^2} - 1}}{{12}}\]

Hence, the correct option for the given problem is D.

Note: In the above question, we have used the formulas of mean and variance. We have also used the properties of arithmetic progression to find the sum of the $n$ natural numbers. The calculations done in the solution are a bit typical and we cannot make mistakes, so do them precisely.. The calculation of mean is simple but it should also be rechecked twice to avoid any mistakes in these types of questions.