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

$\sigma_{ b }^{2}=2 $ (variance of boys)
$n _{1}=$ no. of boys
$\overline{ x }_{ b }=12 $
$n _{2}=$ no. of girls
$\sigma_{ g }^{2}=2$ $\overline{ x }_{ g }=\frac{50 \times 15-12 \times \sigma_{ b }}{30}$
$=\frac{750-12 \times 20}{30}=17=\mu$
variance of combined series
$\sigma^{2}=\frac{n_{1} \sigma_{b}^{2}+n_{2} \sigma_{g}^{2}}{n_{1}+n_{2}}+\frac{n_{1} \cdot n_{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(\bar{x}_{b}-\bar{x}_{g}\right)^{2}$
$\sigma^{2}=\frac{20 \times 2+30 \times 2}{20+30}+\frac{20 \times 30}{(20+30)^{2}}(12-17)^{2}$
$\sigma^{2}=8$
$\Rightarrow \mu+\sigma^{2}=17+8=25$