Paragraph : The mean and standard deviation of 100 items are 50, 5...

Here, $$\begin{aligned} & n_1=100, \quad \bar{x}_1=50, \quad \sigma_1=5 \\ & n_2=150, \quad \bar{x}_2=40, \quad \sigma_2=6 \end{aligned}$$ Combined $\mathrm{SD}=\sigma_{12}=\sqrt{\frac{n_1\left(\sigma_1^2+\mathrm{D}_1^2\right)+n_2\left(\sigma_2^2+\mathrm{D}_2^2\right)}{n_1+n_2}}$$\begin{aligned} \sigma_{12} & =\sqrt{\frac{100\left[5^2+(50-44)^2\right]+}{\frac{150\left[6^2+(40-44)^2\right]}{100+150}}} \\ & =\sqrt{\frac{100[25+36]+150(36+16)}{250}} \end{aligned}$\left[\begin{array}{l} \mathrm{D}_1=\left(\bar{x}_1-\bar{x}_{12}\right) \\ \mathrm{D}_2=\left(\bar{x}_2-\bar{x}_{12}\right) \end{array}\right]\begin{aligned} & =\sqrt{\frac{100 \times 61+150 \times 52}{250}} \\ & =\sqrt{\frac{6100+7800}{250}}=\sqrt{\frac{13900}{250}} \\ & =\sqrt{\frac{1390}{25}}=\sqrt{55.6}=7.45 \cong 7.5 \end{aligned}$$$Hence, combined standard deviation of 250 items$=7.5.