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  • 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 .

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_________.

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x_{C} =\frac{n_{B} \cdot \mu_{B}+n_{G}\: \mu_{G}}{n_{B}+n_{G}} \\

\Rightarrow 15 =\frac{20 \times 12+30 \times \mu}{50} \\

\Rightarrow \mu =\frac{750-240}{30}=17

\sigma_{B}^{2} =\frac{\sum x_{i B}{ }^{2}}{20}-(12)^{2}=2 \\

\Rightarrow \sum_{i B}{ }^{2} =20 \times 146=2920

\sigma_{G}^{2} =\frac{\sum x_{i G^{2}}}{30}-(17)^{2}=2 \\

\Rightarrow \sum_{i a^{2}}=30 \times 291=8730 \\

\sigma^{2} =\frac{\sum x_{i B}\: ^{2}+\sum x_{i G}\: ^{2}}{50}-(15)^{2} 

= \frac{2920+8730}{50}-225

= \frac{11650}{50}-225

= 233-225= 8

So  \mu+\sigma^{2}= 17+8=25

Posted by

Deependra Verma

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