STD 11 - 13. statistics — JEE STD 11 Science — Question

${\sigma ^2} = \frac{{\sum x_1^2}}{n} - {\left( {\overline {.x} } \right)^2}$

$\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}$

${\sigma ^2} = 3434 - 2661 = 833$