Question

The variance of the first 50 even natural numbers is
A) $\dfrac{{833}}{4}$
B) $833$
C) $437$
D) $\dfrac{{437}}{4}$

Answer 1

Here in this question first we have to find the even natural number and then we have to find the variance of the 50 even natural number. Use the formula of variance $\dfrac{{\sum {{x_i}^2} }}{n} - {\left( {\bar x} \right)^2}$. Find the mean of the numbers and the value of $\sum {{x_i}^2}$ to find the variance.

Complete step-by-step answer:
Even natural numbers are
$\Rightarrow 2,4,6,8, \ldots \ldots ,98,100$
The sign of variance is $\sigma$.
The formula of the variance is
${\sigma ^2} = \dfrac{{\sum {{x_i}^2} }}{n} - {\left( {\bar x} \right)^2}$
The mean of the first 50 even natural numbers is,
$\bar x = \dfrac{{\sum {{x_i}} }}{n}$
Here in the formula $x$ represents the series of 50 even natural numbers.
Substitute the values,
$\Rightarrow \bar x = \dfrac{{2 + 4 + 6 + \ldots + 100}}{{50}}$
Take 2 commons from the sum,
$\Rightarrow \bar x = 2 \times \dfrac{{1 + 2 + 3 + \ldots + 50}}{{50}}$
The formula for the sum of natural number is,
$\dfrac{{n\left( {n + 1} \right)}}{2}$
Use the formula with $n = 50$,
$\Rightarrow \bar x = \dfrac{2}{{50}} \times \dfrac{{50\left( {50 + 1} \right)}}{2}$
Simplify the terms,
$\Rightarrow \bar x = 51$
Now find the sum of square on 50 even natural terms,
$\sum {{x_i}^2} = {2^2} + {4^2} + {6^2} + \ldots + {100^2}$
Take ${2^2}$ common from the sum,
$\Rightarrow \sum {{x_i}^2} = {2^2} \times \dfrac{{{1^2} + {2^2} + {3^2} + \ldots + {{50}^2}}}{{50}}$
The formula for the sum of the square of natural number is,
$\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
Use the formula with $n = 50$,
$\Rightarrow \sum {{x_i}^2} = \dfrac{4}{{50}} \times \dfrac{{50\left( {50 + 1} \right)\left( {100 + 1} \right)}}{6}$
Simplify the terms,
$\Rightarrow \sum {{x_i}^2} = 2 \times 17 \times 101$
Multiply the terms,
$\Rightarrow \sum {{x_i}^2} = 3434$
Substitute the values in the formula,
$\Rightarrow {\sigma ^2} = 3434 - {\left( {51} \right)^2}$
Square the terms,
$\Rightarrow {\sigma ^2} = 3434 - 2601$
Subtract the terms,
$\therefore {\sigma ^2} = 833$
So, the variance is 833,

Hence, option (B) is the correct answer.

Note: Here in this type of question students get confused in the calculation parts of variance. Sometimes students take the series of 50 even natural numbers as arithmetic progression and apply the formula of the sum of 50 even natural numbers instead of doing variance. Solve the question step by step for getting the correct answer. In the second part of the variance there are squares, so do not forget to do the square at the end of the solution of the second part.