
The variance of the first 50 even natural numbers is?
Answer
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Hint: 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. Separate the part of the variance for solving your question easily.
Complete step-by-step answer:
Even natural numbers are
2, 4, 6, 8, . . . . . . . . ., 98, 100
The sigh of variance is $\sigma $.
The formula of the variance is
${\sigma ^2} = \dfrac{{\sum {{x_1}^2} }}{n} - {(\bar x)^2}$
Here in the formula x represent the series of 50 even natural numbers
Here x is equal to the series of 50 even natural numbers
${\sigma ^2} = \dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}} - {(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$ ……….. (1)
Separating the above equation, we get two parts which is
${i_1}$= $\dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}}$
${i_2}$= ${(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$
So, the
${\sigma ^2} = {i_1} - {i_2}$
Separate the variance and Solve
${i_1}$= $\dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}}$
Takin ${2^2}$as common
${i_1} = {2^2}\dfrac{{{1^2} + {2^2} + {3^2} + ....... + {{50}^2}}}{{50}}$
Here sum of square of natural number
$n(n + 1)(2n + 1)$, $n = 50$
Put the values in formula
${i^2} = \dfrac{{{2^2}}}{{50}}.50(50 + 1)(100 + 1)$
$ = 3434$
Solving the ${i_2}$
${i_2}$= ${(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$
Here sum of natural number is
$\dfrac{{}}{{}}$$\dfrac{{n(n + 1)}}{2}$
$ = {(\dfrac{{50.\dfrac{{2 + 100}}{2}}}{{50}})^2}$
Solving the above equation, we get
$ = {(51)^2}$
$ = 2661$
Putting the values in the equation (1)
${\sigma ^2} = 3434 - 2661$
Subtract the above equation
${\sigma ^2} = 833$
Hence the variance of the 50 even natural numbers is 833.
Note: Here in this type of question students get confused in the calculation parts of variance. Sometimes students take the series of 50 even natural numbers as arithmetic progression and apply the formula of sum of 50 even natural numbers instead of doing variance. Solve the question step by step for getting the correct answer. In the second part of the variance there are squares, so do not forget to do the square at the end of the solution of the second part.
Complete step-by-step answer:
Even natural numbers are
2, 4, 6, 8, . . . . . . . . ., 98, 100
The sigh of variance is $\sigma $.
The formula of the variance is
${\sigma ^2} = \dfrac{{\sum {{x_1}^2} }}{n} - {(\bar x)^2}$
Here in the formula x represent the series of 50 even natural numbers
Here x is equal to the series of 50 even natural numbers
${\sigma ^2} = \dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}} - {(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$ ……….. (1)
Separating the above equation, we get two parts which is
${i_1}$= $\dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}}$
${i_2}$= ${(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$
So, the
${\sigma ^2} = {i_1} - {i_2}$
Separate the variance and Solve
${i_1}$= $\dfrac{{{2^2} + {4^2} + {6^2} + ........ + {{100}^2}}}{{50}}$
Takin ${2^2}$as common
${i_1} = {2^2}\dfrac{{{1^2} + {2^2} + {3^2} + ....... + {{50}^2}}}{{50}}$
Here sum of square of natural number
$n(n + 1)(2n + 1)$, $n = 50$
Put the values in formula
${i^2} = \dfrac{{{2^2}}}{{50}}.50(50 + 1)(100 + 1)$
$ = 3434$
Solving the ${i_2}$
${i_2}$= ${(\dfrac{{2 + 4 + 6 + ...... + 100}}{{50}})^2}$
Here sum of natural number is
$\dfrac{{}}{{}}$$\dfrac{{n(n + 1)}}{2}$
$ = {(\dfrac{{50.\dfrac{{2 + 100}}{2}}}{{50}})^2}$
Solving the above equation, we get
$ = {(51)^2}$
$ = 2661$
Putting the values in the equation (1)
${\sigma ^2} = 3434 - 2661$
Subtract the above equation
${\sigma ^2} = 833$
Hence the variance of the 50 even natural numbers is 833.
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 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.
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