1. Problem: Given $X \sim \text{Poisson}(50)$, use the normal approximation to find: a) $P(52 < X < 56)$ b) $P(50 \leq X \leq 52)$ 2. The Poisson distribution with parameter $\lambda=50$ can be approximated by a normal distribution $N(\mu=50, \sigma^2=50)$ because $\lambda$ is large. 3. For the normal approximation, use continuity correction: - a) $P(52 < X < 56) = P(52.5 \leq X \leq 55.5)$ - b) $P(50 \leq X \leq 52) = P(49.5 \leq X \leq 52.5)$ 4. Standardize using $Z = \frac{X - \mu}{\sigma}$ where $\sigma = \sqrt{50} \approx 7.071$: - a) $P(52.5 \leq X \leq 55.5) = P\left(\frac{52.5 - 50}{7.071} \leq Z \leq \frac{55.5 - 50}{7.071}\right) = P(0.354 \leq Z \leq 0.777)$ - b) $P(49.5 \leq X \leq 52.5) = P\left(\frac{49.5 - 50}{7.071} \leq Z \leq \frac{52.5 - 50}{7.071}\right) = P(-0.071 \leq Z \leq 0.354)$ 5. Use standard normal tables or a calculator: - a) $P(0.354 \leq Z \leq 0.777) = \Phi(0.777) - \Phi(0.354) \approx 0.781 - 0.638 = 0.143$ - b) $P(-0.071 \leq Z \leq 0.354) = \Phi(0.354) - \Phi(-0.071) \approx 0.638 - 0.471 = 0.167$ --- 6. Problem: In 1 ml pond water, average microorganisms = 117. For 10 ml, mean $\lambda = 1170$. Find $P(X > 1200)$ using normal approximation. 7. Approximate $X \sim \text{Poisson}(1170)$ by $N(1170, 1170)$. 8. Use continuity correction: $P(X > 1200) = P(X \geq 1201) \approx P(Y \geq 1200.5)$ where $Y \sim N(1170, 1170)$. 9. Standardize: $$Z = \frac{1200.5 - 1170}{\sqrt{1170}} = \frac{30.5}{34.21} \approx 0.892$$ 10. Find $P(Z > 0.892) = 1 - \Phi(0.892) \approx 1 - 0.813 = 0.187$. --- 11. Problem: Delayed trains occur at rate 8 per day. For 14 days, $\lambda = 8 \times 14 = 112$. a) Find $P(X < 100)$ where $X \sim \text{Poisson}(112)$ using normal approximation. b) Decide if Poisson model is reasonable. 12. Approximate $X \sim N(112, 112)$. 13. Use continuity correction: $P(X < 100) = P(X \leq 99) \approx P(Y \leq 99.5)$. 14. Standardize: $$Z = \frac{99.5 - 112}{\sqrt{112}} = \frac{-12.5}{10.58} \approx -1.18$$ 15. Find $P(Z < -1.18) = \Phi(-1.18) \approx 0.119$. 16. Poisson model is reasonable if events are independent and occur at a constant average rate. Since delayed trains are rare and independent daily, Poisson is appropriate. --- 17. Problem: Fabio receives on average 48 emails/day, can answer at most 60. Find $P(X \leq 60)$ where $X \sim \text{Poisson}(48)$. 18. Approximate $X \sim N(48, 48)$. 19. Use continuity correction: $P(X \leq 60) \approx P(Y \leq 60.5)$. 20. Standardize: $$Z = \frac{60.5 - 48}{\sqrt{48}} = \frac{12.5}{6.93} \approx 1.80$$ 21. Find $P(Z \leq 1.80) = \Phi(1.80) \approx 0.964$. --- 22. Problem: $X \sim \text{Poisson}(42)$, find minimum integer $x$ such that $P(X > x) \leq 0.1$. 23. Use normal approximation $N(42, 42)$. 24. We want $P(X > x) = 1 - P(X \leq x) \leq 0.1 \Rightarrow P(X \leq x) \geq 0.9$. 25. Find $z$ such that $\Phi(z) = 0.9$, $z \approx 1.28$. 26. Use continuity correction: $P(X \leq x) \approx P(Y \leq x + 0.5)$. 27. Solve for $x$: $$\frac{x + 0.5 - 42}{\sqrt{42}} = 1.28 \Rightarrow x + 0.5 = 42 + 1.28 \times 6.48 = 42 + 8.29 = 50.29$$ 28. Minimum integer $x = 50$. Final answers: a) $P(52 < X < 56) \approx 0.143$ b) $P(50 \leq X \leq 52) \approx 0.167$ 3) $P(X > 1200) \approx 0.187$ 4a) $P(X < 100) \approx 0.119$ 4b) Poisson model reasonable due to independence and constant rate. 5) $P(X \leq 60) \approx 0.964$ 6) Minimum $x = 50$ for $P(X > x) \leq 0.1$