Question

The variance of the first \[50\] even natural number is?

Answer 1

Hint: 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.