Question

Variance of first $20$ natural number is
A. $\dfrac{133}{4}$
B. $\dfrac{379}{12}$
C. $\dfrac{133}{2}$
D. $\dfrac{399}{4}$

Answer 1

Hint: Here we have to find the variance of first $20$ natural numbers. Firstly we will write the formula for calculating the variance then we will substitute the values in the formula by taking the number of observations as $n$ . Finally we will simplify the equation and put $n=20$ in it and solve it to get our desired answer.

Complete step-by-step solution:
We have to find the variance of the first $20$ numbers.
As we know that variance is calculated as follows,
$\sigma =\dfrac{\sum{{{\left( {{x}_{i}} \right)}^{2}}}}{n}-{{\mu }^{2}}$….$\left( 1 \right)$
Where, \[\sigma =\] Variance and $\mu =$ Mean
We know the mean of natural number is as follows,
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 $n=20$ ,
$\Rightarrow \sigma =\dfrac{{{20}^{2}}-1}{12}$
$\Rightarrow \sigma =\dfrac{400-1}{12}$
So we get,
$\Rightarrow \sigma =\dfrac{399}{12}$
$\Rightarrow \sigma =\dfrac{133}{4}$
So we get the variance of the first $20$ natural number as $\dfrac{133}{4}$ .
Hence correct option is (A).

Note: Variance is the expectation of the square deviation of a random variable from its mean that means we subtract the deviation by the mean to get the variation for a given data. It is used to measure how far away the data is scattered out from the average value of the observation. It has a very important and central role in statistics. It is also calculated as square of the standard deviation. It is the second central moment of distribution and the covariance of the random variable with itself.