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The variance of first $50$ even natural numbers is
$437$
$\frac{{437}}{4}$
$\frac{{833}}{4}$
$833$
Solution
$2,4,6,8,……,98,100$
${\sigma ^2} = \frac{{\sum x_1^2}}{n} – {\left( {\overline {.x} } \right)^2}$
$\frac{{{2^2} + {4^2} + {6^2} + …. + {{100}^2}}}{{50}}$$ – {\left( {\frac{{2 + 4 + 6 + …. + 100}}{{50}}} \right)^2}$
${i_1} = \frac{{{2^2} + {4^2} + {6^2} + …. + {{100}^2}}}{{50}}$
$ = {2^2}\frac{{{1^2} + {2^2} + {3^2} + … + {{50}^2}}}{{50}}$
$ = \frac{{{2^2}}}{{50}} \times 50\left( {50 + 1} \right)\left( {100 + 1} \right)$
$ = 3434$
${i_2} = {\left( {\frac{{2 + 4 + 6 + ….. + 100}}{{50}}} \right)^2}$
$ = {\left( {\frac{{50 \times \frac{{2 + 100}}{2}}}{{50}}} \right)^2}$
$ = {\left( {51} \right)^2}$
${\sigma ^2} = 3434 – 2661 = 833$
Similar Questions
If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is
| $X$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
| $f$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$ |
