Question

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 1

Sum of marks of boys, $\sum X_B=240$
Total marks $\Rightarrow \sum X=750$
So, sum of marks of girls $=510=\sum X_G$

$\Rightarrow \frac{\sum X_B^2}{20}-(12{)}^2=2\text{and}\frac{\sum X_G^2}{30}-({\overline{X}}_G{)}^2=2$
$\Rightarrow \sum X_B^2=2920\text{and}\frac{\sum X_G^2}{30}-(17{)}^2=2$
$\therefore \sum X_G^2=8730$

$\text{Overall Variance}=\frac{\sum X_B^2+X_G^2}{50}-(\overline{X}{)}^2$
$=\frac{2920+8730}{50}-(15{)}^2=8$
$\Rightarrow \mu =17,{\sigma }^2=8$
$\therefore \mu +{\sigma }^2=25$