1. **Problem Statement:** We have two groups of workers with given means and standard deviations of their monthly wages. We need to find the standard deviation of the combined group. 2. **Given Data:** - Group 1 (Male workers): $n_1 = 50$, mean $\bar{x}_1 = 630$, standard deviation $s_1 = 90$ - Group 2 (Female workers): $n_2 = 40$, mean $\bar{x}_2 = 540$, standard deviation $s_2 = 60$ 3. **Formula for combined standard deviation:** $$s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + \frac{n_1 n_2}{n_1 + n_2} (\bar{x}_1 - \bar{x}_2)^2}{n_1 + n_2 - 1}}$$ This formula accounts for the variance within each group and the variance between the group means. 4. **Calculate each term:** - $(n_1 - 1)s_1^2 = 49 \times 90^2 = 49 \times 8100 = 396900$ - $(n_2 - 1)s_2^2 = 39 \times 60^2 = 39 \times 3600 = 140400$ - Difference of means squared: $(630 - 540)^2 = 90^2 = 8100$ - Weighted mean difference term: $\frac{50 \times 40}{50 + 40} \times 8100 = \frac{2000}{90} \times 8100 = 22.222... \times 8100 = 180000$ 5. **Sum numerator:** $$396900 + 140400 + 180000 = 717300$$ 6. **Denominator:** $$50 + 40 - 1 = 89$$ 7. **Calculate combined variance:** $$\frac{717300}{89} \approx 8058.43$$ 8. **Standard deviation:** $$s = \sqrt{8058.43} \approx 89.78$$ **Final answer:** The standard deviation of monthly wages for the combined group is approximately $89.78$.