$\therefore n_{2}=250-100 \Rightarrow n_{2}=150$
$\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$
$15.6=\frac{100(15)+(150)\left(\bar{x}_{2}\right)}{250}$
$\Rightarrow \overline{\mathrm{x}}_{2}=16$
$\overline{\mathrm{X}}_{1}=15 \quad\quad\quad\quad \Rightarrow$
$\sigma_{1}^{2}=V_{1}(x)=9 \quad \sigma^{2}=\operatorname{Var}(x)=13.44$
$\sigma^{2}=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}$
$\mathrm{n}_{2}=150, \overline{\mathrm{x}}_{2}=16, \mathrm{~V}_{2}(\mathrm{x})=\sigma_{2}$
$13.44=\frac{100 \times 9+150 \times \sigma_{2}^{2}}{250}+\frac{100 \times 150}{(250)^{2}} \times 1$
$\Rightarrow \sigma_{2}^{2}=16 \Rightarrow \sigma_{2}=4$
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