1. **Problem statement:** Given a table of workers by gender and company, find various probabilities related to company and gender. 2. **Total workers:** $240$. 3. **(i) Probability worker works in company Z:** Total workers in company Z = $75$. $$P(Z) = \frac{75}{240} = \frac{5}{16} = 0.3125$$ 4. **(ii) Probability worker is male and works in company Y:** Male workers in company Y = $35$. $$P(M \cap Y) = \frac{35}{240} = \frac{7}{48} \approx 0.1458$$ 5. **(iii) Probability worker is male or works in company X:** Male workers total = $115$. Workers in company X total = $100$. Male workers in company X = $55$. Use formula: $$P(M \cup X) = P(M) + P(X) - P(M \cap X)$$ $$= \frac{115}{240} + \frac{100}{240} - \frac{55}{240} = \frac{160}{240} = \frac{2}{3} \approx 0.6667$$ 6. **(iv) Probability worker works in company Z given female:** Female workers total = $125$. Female workers in company Z = $50$. Conditional probability: $$P(Z|F) = \frac{P(Z \cap F)}{P(F)} = \frac{50/240}{125/240} = \frac{50}{125} = \frac{2}{5} = 0.4$$ --- 7. **Binomial probability problem:** Success probability $p=0.08$, number of trials $n=10$. 8. **(i) Probability exactly 5 students succeed:** Use binomial formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ $$P(X=5) = \binom{10}{5} (0.08)^5 (0.92)^5$$ Calculate: $$\binom{10}{5} = 252$$ $$P(X=5) = 252 \times (0.08)^5 \times (0.92)^5$$ 9. **(ii) Probability at least 2 succeed:** $$P(X \geq 2) = 1 - P(X<2) = 1 - [P(X=0) + P(X=1)]$$ Calculate: $$P(X=0) = \binom{10}{0} (0.08)^0 (0.92)^{10} = (0.92)^{10}$$ $$P(X=1) = \binom{10}{1} (0.08)^1 (0.92)^9 = 10 \times 0.08 \times (0.92)^9$$ --- 10. **Normal distribution problem:** Mean $\mu=80$, standard deviation $\sigma=23$. 11. **(i) Probability time less than 60 minutes:** Calculate z-score: $$z = \frac{60 - 80}{23} = \frac{-20}{23} \approx -0.8696$$ Use standard normal table or calculator: $$P(X<60) = P(Z < -0.8696) \approx 0.1922$$ 12. **(ii) Probability time between 70 and 110 minutes:** Calculate z-scores: $$z_1 = \frac{70 - 80}{23} = -\frac{10}{23} \approx -0.4348$$ $$z_2 = \frac{110 - 80}{23} = \frac{30}{23} \approx 1.3043$$ Probability: $$P(70 < X < 110) = P(z_1 < Z < z_2) = P(Z < 1.3043) - P(Z < -0.4348)$$ From standard normal table: $$\approx 0.9032 - 0.3320 = 0.5712$$