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QJEEM21Ma65ca: JEE Main 2021 Mathematics - Class 11 JEEM21Ma65ca Solution: An online exam is attempted by 50 candidates out of which 20…

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 μ\mu is the average marks of girls and σ\sigma 2 is the variance of marks of 50 candidates, then μ\mu + σ\sigma 2 is equal to ________________.

Step-by-Step Verified Solution

σb2\sigma _b^2 = 2 (variance of boys)

n1 = no. of boys

x‾b{\overline x _b} = 12

n2 = no. of girls

σg2\sigma _g^2 = 2

x‾g{\overline x _g} = 50×15−12×σb30=750−12×2030=17=μ{{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu

variance of combined series

σ2=n1σb2+n2σg2n1+n2+n1. n2(n1+n2)2(x‾b−x‾g)2\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}

σ2=20×2+30×220+30+20×30(20+30)2(12−17)2\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}

σ\sigma 2 = 8

⇒\Rightarrow μ\mu + σ\sigma 2 = 17 + 8 = 25

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