Question

The variance of the first $50$ even natural number is?

Answer 1

The even numbers are start with $2$ and end with $100$
Use a variance formula for $n$ natural numbers because there is no individual formula for $n$ even natural numbers, by simplifying the even natural number or by taking out common terms in that, we get the $n$ natural number form.
Substitute it, then also sometime we may use the sum of $n$ natural numbers also, the sum of ${n^2}$ natural numbers.
Hence, we get the final answer.

Formula used: The variance of $n$ natural number is ${\sigma ^2} = \dfrac{{\sum {{{({x_i})}^2}} }}{n} - {\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)^2}$
The sum of square of first n natural number is $\dfrac{{n(n + 1)(2n + 1)}}{6}$
The sum of the first n natural number is $\dfrac{{n(n + 1)}}{2}$

Complete step-by-step answer:
It is given that the set of $50$ even numbers is $2,4,6,8,10,12,.......90,92,94,96,98,100$
The formula for variance of $n$ natural number is
${\sigma ^2} = \dfrac{{\sum {{{({x_i})}^2}} }}{n} - {\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)^2}....\left( 1 \right)$
Given, $n = 50$,
$\sum {{{({x_i})}^2}} = {2^2} + {4^2} + {6^2} + {8^2} + ...... + {98^2} + {100^2}$, that is the sum of the square of the first $50$ even natural number.
The sum of the first $50$ even natural number, we can write it as,
$\sum {{x_i} = 2 + 4 + 6 + 8 + ..... + 100}$
Substitute the terms in equation $\left( 1 \right)$ we get,
${\sigma ^2} = \dfrac{{\sum {{{({x_i})}^2}} }}{{50}} - {\left( {\dfrac{{\sum {{x_i}} }}{{50}}} \right)^2}....\left( 2 \right)$
Here we solve one by one,
First we take,
$\sum {\dfrac{{{{({x_i})}^2}}}{{50}}} = \dfrac{{{2^2} + {4^2} + {6^2} + {8^2} + ...... + {{98}^2} + {{100}^2}}}{{50}}....\left( 3 \right)$,
We have to write it as rearrange the RHS, we get,

$$2 = 1 \times 2 \\\ 4 = 2 \times 2 \\\ 6 = 2 \times 3 \\\ . \\\ . \\\ . \\\ 98 = 2 \times 49 \\\ 100 = 2 \times 50 \\\$$

Substitute the above rearranged terms in equation $\left( 3 \right)$ we get,
$\sum {\dfrac{{{{({x_i})}^2}}}{{50}}} = \dfrac{{{{(1 \times 2)}^2} + {{\left( {2 \times 2} \right)}^2} + {{\left( {2 \times 3} \right)}^2} + ....... + {{\left( {2 \times 49} \right)}^2} + {{\left( {2 \times 50} \right)}^2}}}{{50}}$
Partially split the squares,
$\sum {\dfrac{{{{({x_i})}^2}}}{{50}}} = \dfrac{{{1^2} \times {2^2} + {2^2} \times {2^2} + {2^2} \times {3^2} + ....... + {2^2} \times {{49}^2} + {2^2} \times {{50}^2}}}{{50}}$
Take ${2^2}$ has common in all terms,
$\sum {\dfrac{{{{({x_i})}^2}}}{{50}}} = \dfrac{{{2^2}({1^2} + {2^2} + {3^2} + ....... + {{48}^2} + {{50}^2})}}{{50}}$
On squaring the term, we get
$\sum {\dfrac{{{{({x_i})}^2}}}{{50}}} = \dfrac{{4({1^2} + {2^2} + {3^2} + ....... + {{48}^2} + {{50}^2})}}{{50}}$
$\dfrac{{\sum {{{({x_i})}^2}} }}{n} = \dfrac{{4({1^2} + {2^2} + {3^2} + ....... + {{48}^2} + {{50}^2})}}{{50}}$
On dividing the terms we get,
$\dfrac{{\sum {{{({x_i})}^2}} }}{n} = \dfrac{{2({1^2} + {2^2} + {3^2} + ....... + {{48}^2} + {{50}^2})}}{{25}}....\left( 4 \right)$
Now, we have to find
$({1^2} + {2^2} + {3^2} + ....... + {48^2} + {50^2})$
Here we use the formula for sum of the square of first ${n^2}$,
$\dfrac{{n(n + 1)(2n + 1)}}{6}$
Here$n = 50$,
$\dfrac{{50(50 + 1)(2 \times 50 + 1)}}{6}$
On adding and multiplying the terms
$\dfrac{{50(51)(100 + 1)}}{6}$
On dividing the first terms with denominator
$\dfrac{{25(51)(100 + 1)}}{3}$
On dividing the second term with denominator,
$25 \times 17 \times 101$
On multiplying the terms we get, $42925$
$({1^2} + {2^2} + {3^2} + ....... + {48^2} + {50^2}) = 42925$
Substitute $({1^2} + {2^2} + {3^2} + ....... + {48^2} + {50^2})$=$42925$ in $\left( 4 \right)$ we get,
$\dfrac{{\sum {{{({x_i})}^2}} }}{n} = \dfrac{{2\left( {42925} \right)}}{{25}}$
On dividing the terms we get,
$\dfrac{{\sum {{{({x_i})}^2}} }}{n} = 2 \times 1717$
Let us multiply,
$\dfrac{{\sum {{{({x_i})}^2}} }}{n} = 3434....\left( 5 \right)$
From $\left( 2 \right)$ we take,
$\dfrac{{\sum {{x_i}} }}{n} = \dfrac{{2 + 4 + 6 + .... + 100}}{{50}}....\left( 6 \right)$
Take
$\sum {{x_i} = 2 + 4 + 6 + ... + 100}$
We have to rearrange the RHS we get,
$\sum {{x_i} = 2 \times 1} + 2 \times 2 + 2 \times 3 + .... + 2 \times 50$
Take $2$ has common in all terms
$\sum {{x_i} = 2(1} + 2 + 3 + .... + 50)$
Now, we have to find $(1 + 2 + 3..... + 50)$
Here we use the formula for the sum first $n$ natural numbers,
$\dfrac{{n(n + 1)}}{2}$
Here $n = 50$
$\dfrac{{50(50 + 1)}}{2}$
On adding we get,
$\dfrac{{50 \times 51}}{2}$
Let us divided the first number with the denominator we get,
$25 \times 51$
On multiply
$1275$
$(1 + 2 + 3..... + 50)$ = $1275$
Substitute $(1 + 2 + 3..... + 50)$=$1275$ in $\left( 6 \right)$
$\sum {\dfrac{{{x_i}}}{{50}} = \dfrac{{2\left( {1275} \right)}}{{50}}}$
On multiplying the numerator term we get
$\dfrac{{\sum {{x_i}} }}{n} = \dfrac{{2550}}{{50}}$
On dividing,
$\dfrac{{\sum {{x_i}} }}{n} = 51$
Taking square on both sides we get,
${\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)^2} = {(51)^2}$
On squaring the LHS,
${\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)^2} = 2601....\left( 7 \right)$
Substitute $\dfrac{{\sum {{{({x_i})}^2}} }}{n} = 3434$ and ${\left( {\dfrac{{\sum {{x_i}} }}{n}} \right)^2} = 2601$ in $\left( 1 \right)$we get,
${\sigma ^2} = 3434 - 2601$
On subtracting
${\sigma ^2} = 833$
Hence, the variance of the first $50$ even natural number is $833$

Note: There is no individual formula for $n$ even natural number, so that we use $n$ natural number formula,
Followed by small simplification, we get the form to $n$ natural even number.
If we take square root for variance we get standard deviation for $50$ even natural numbers.