JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 18)
An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If $$\mu$$ is the average marks of girls and $$\sigma$$2 is the variance of marks of 50 candidates, then $$\mu$$ + $$\sigma$$2 is equal to ________________.
Answer
25
Explanation
$$\sigma _b^2$$ = 2 (variance of boys)
n1 = no. of boys
$${\overline x _b}$$ = 12
n2 = no. of girls
$$\sigma _g^2$$ = 2
$${\overline x _g}$$ = $${{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu $$
variance of combined series
$$\sigma _{}^2 = {{{n_1}\sigma _b^2 + {n_2}\sigma _g^2} \over {{n_1} + {n_2}}} + {{{n_1}.\,{n_2}} \over {{{({n_1} + {n_2})}^2}}}{\left( {{{\overline x }_b} - {{\overline x }_g}} \right)^2}$$
$$\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}$$
$$\sigma$$2 = 8
$$\Rightarrow$$ $$\mu$$ + $$\sigma$$2 = 17 + 8 = 25
n1 = no. of boys
$${\overline x _b}$$ = 12
n2 = no. of girls
$$\sigma _g^2$$ = 2
$${\overline x _g}$$ = $${{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu $$
variance of combined series
$$\sigma _{}^2 = {{{n_1}\sigma _b^2 + {n_2}\sigma _g^2} \over {{n_1} + {n_2}}} + {{{n_1}.\,{n_2}} \over {{{({n_1} + {n_2})}^2}}}{\left( {{{\overline x }_b} - {{\overline x }_g}} \right)^2}$$
$$\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}$$
$$\sigma$$2 = 8
$$\Rightarrow$$ $$\mu$$ + $$\sigma$$2 = 17 + 8 = 25
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