The variance of first 50 even natural numbers is

Question

Accepted Answer

Here is total $50$ numbers, so $N=50$

Variance $=$ $${{\sum {{x^2}} } \over {50}} - {\left( {{{\sum x } \over {50}}} \right)^2}$$

Here $$\sum {{x^2}}  = $$ sum of square of first $50$ even natural number.

$= {2^2} + {4^2} + ..... + {100^2}$

$$ = {2^2}\left[ {{1^2} + {2^2} + ....... + {{50}^2}} \right]$$

$$ = 4\left[ {{{50 \times 51 \times 101} \over 6}} \right]$$

So, $${{\sum {{x^2}} } \over {50}} = {{4 \times 51 \times 101} \over 6} = 3434$$

$$\sum {x = } $$ sum of first $50$ even natural numbers

$= 2 + 4 + ...... + 100$

$$ = 2\left[ {1 + 2 + .... + 50} \right]$$

$$ = 2\left[ {{{50 \times 51} \over 2}} \right]$$

$$ = 50 \times 51$$

$$\therefore\,\,\,$$ $${\left( {{{\sum x } \over {50}}} \right)^2} = {\left( {{{50 \times 51} \over {50}}} \right)^2} = 2601$$

$$\therefore\,\,\,$$ Variance $= 3434 - 2601$ $=833$

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