1. **Problem Statement:** We have two parts: (a) a binomial probability problem about attendance at a stadium where 70% are males, and (b) a normal distribution problem about meals served at a café with mean 6000 and standard deviation 600. --- ### Part (a): Binomial Probability 2. **Given:** - Probability of male spectator $p=0.7$ - Number of trials $n=7$ 3. **Formulas:** - Binomial probability: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - For "at most" probabilities: $$P(X \leq k) = \sum_{i=0}^k P(X=i)$$ 4. **(i) Probability exactly 4 males:** - Calculate $P(X=4)$ where $X$ = number of males. - $$P(X=4) = \binom{7}{4} (0.7)^4 (0.3)^3$$ - Calculate combination: $\binom{7}{4} = \frac{7!}{4!3!} = 35$ - Calculate powers: $(0.7)^4 = 0.2401$, $(0.3)^3 = 0.027$ - Multiply: $35 \times 0.2401 \times 0.027 = 35 \times 0.0064827 = 0.2269$ 5. **(ii) Probability at most 5 females:** - Females probability $q=0.3$ - Number of females $Y$ in 7 people is binomial with $p=0.3$ - "At most 5 females" means $P(Y \leq 5) = 1 - P(Y=6) - P(Y=7)$ - Calculate $P(Y=6)$: $$P(Y=6) = \binom{7}{6} (0.3)^6 (0.7)^1 = 7 \times 0.000729 \times 0.7 = 0.00357$$ - Calculate $P(Y=7)$: $$P(Y=7) = \binom{7}{7} (0.3)^7 (0.7)^0 = 1 \times 0.0002187 \times 1 = 0.0002187$$ - Sum: $0.00357 + 0.0002187 = 0.00379$ - So, $P(Y \leq 5) = 1 - 0.00379 = 0.9962$ --- ### Part (b): Normal Distribution 6. **Given:** - Mean $\mu=6000$ - Standard deviation $\sigma=600$ 7. **Standardize using z-score:** $$z = \frac{X - \mu}{\sigma}$$ 8. **(i) Probability number of meals $\leq 5000$:** - Calculate $z$: $$z = \frac{5000 - 6000}{600} = \frac{-1000}{600} = -1.6667$$ - Use standard normal table or calculator: $$P(Z \leq -1.6667) \approx 0.0478$$ 9. **(ii) Probability number of meals $> 7500$:** - Calculate $z$: $$z = \frac{7500 - 6000}{600} = \frac{1500}{600} = 2.5$$ - Probability: $$P(Z > 2.5) = 1 - P(Z \leq 2.5) = 1 - 0.9938 = 0.0062$$ 10. **(iii) Probability between 5500 and 6500:** - Calculate $z$ for 5500: $$z_1 = \frac{5500 - 6000}{600} = -0.8333$$ - Calculate $z$ for 6500: $$z_2 = \frac{6500 - 6000}{600} = 0.8333$$ - Probability: $$P(5500 \leq X \leq 6500) = P(-0.8333 \leq Z \leq 0.8333) = P(Z \leq 0.8333) - P(Z \leq -0.8333)$$ - From tables: $$P(Z \leq 0.8333) \approx 0.7977, \quad P(Z \leq -0.8333) \approx 0.2023$$ - So: $$0.7977 - 0.2023 = 0.5954$$ 11. **(iv) Find $x$ such that $P(X < x) = 0.4$:** - Find $z$ for 0.4 quantile: $$z \approx -0.253$$ - Convert back: $$x = \mu + z \sigma = 6000 + (-0.253)(600) = 6000 - 151.8 = 5848.2$$ 12. **(v) Find $x$ such that $P(X > x) = 0.2$ (top 20%):** - So $P(X \leq x) = 0.8$ - Find $z$ for 0.8 quantile: $$z \approx 0.8416$$ - Convert back: $$x = 6000 + 0.8416 \times 600 = 6000 + 504.96 = 6504.96$$ --- **Final answers:** - (a)(i) $P(4 \text{ males}) = 0.2269$ - (a)(ii) $P(\leq 5 \text{ females}) = 0.9962$ - (b)(i) $P(X \leq 5000) = 0.0478$ - (b)(ii) $P(X > 7500) = 0.0062$ - (b)(iii) $P(5500 \leq X \leq 6500) = 0.5954$ - (b)(iv) $x = 5848.2$ - (b)(v) $x = 6504.96$