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