1. **State the problem:** We have 30 people: 20 men and 10 women. Men's mean weight $\mu_m = 65$ kg, variance $\sigma_m^2 = 56.25$ kg$^2$. Women's mean weight $\mu_w = 51.3$ kg, variance $\sigma_w^2 = 24.01$ kg$^2$. We want the standard deviation of the combined group. 2. **Recall formulas:** - Variance $\sigma^2 = E(X^2) - (E(X))^2$. - For combined groups, total mean: $$\mu = \frac{n_m \mu_m + n_w \mu_w}{n_m + n_w}$$ - Total variance: $$\sigma^2 = \frac{1}{n_m + n_w} \left( n_m (\sigma_m^2 + \mu_m^2) + n_w (\sigma_w^2 + \mu_w^2) \right) - \mu^2$$ 3. **Calculate total mean:** $$\mu = \frac{20 \times 65 + 10 \times 51.3}{30} = \frac{1300 + 513}{30} = \frac{1813}{30} = 60.4333$$ 4. **Calculate combined variance:** Calculate $n_m (\sigma_m^2 + \mu_m^2)$: $$20 \times (56.25 + 65^2) = 20 \times (56.25 + 4225) = 20 \times 4281.25 = 85625$$ Calculate $n_w (\sigma_w^2 + \mu_w^2)$: $$10 \times (24.01 + 51.3^2) = 10 \times (24.01 + 2631.69) = 10 \times 2655.7 = 26557$$ Sum: $$85625 + 26557 = 112182$$ Divide by total number: $$\frac{112182}{30} = 3739.4$$ Subtract square of total mean: $$3739.4 - (60.4333)^2 = 3739.4 - 3652.2 = 87.2$$ 5. **Calculate standard deviation:** $$\sigma = \sqrt{87.2} = 9.34$$ **Final answer:** The standard deviation of the weights of all 30 people is approximately **9.34 kg**.