1. **Problem:** The random variable $X$ is normally distributed with standard deviation 25. Given $P(X < 27.5) = 0.3085$, find the mean $\mu$. 2. **Formula:** For a normal distribution, the standardized variable $Z = \frac{X - \mu}{\sigma}$ follows the standard normal distribution $N(0,1)$. 3. **Step:** Find the $z$-score corresponding to $P(Z < z) = 0.3085$. From standard normal tables, $z \approx -0.5$. 4. **Step:** Use $z = \frac{27.5 - \mu}{25} = -0.5$. 5. **Step:** Solve for $\mu$: $$27.5 - \mu = -0.5 \times 25$$ $$27.5 - \mu = -12.5$$ $$\mu = 27.5 + 12.5 = 40$$ --- 6. **Problem:** $X \sim N(45, \sigma)$, and $P(X > 51) = 0.288$. Find $\sigma$. 7. **Step:** Find $z$ such that $P(Z > z) = 0.288$, so $P(Z < z) = 0.712$. From tables, $z \approx 0.56$. 8. **Step:** Use $z = \frac{51 - 45}{\sigma} = 0.56$. 9. **Step:** Solve for $\sigma$: $$\sigma = \frac{51 - 45}{0.56} = \frac{6}{0.56} \approx 10.71$$ --- 10. **Problem:** Volume $X \sim N(333, \sigma)$, with $P(X > 340) = 0.20$. 11. **Step:** Find $z$ such that $P(Z > z) = 0.20$, so $P(Z < z) = 0.80$. From tables, $z \approx 0.84$. 12. **Step:** Use $z = \frac{340 - 333}{\sigma} = 0.84$. 13. **Step:** Solve for $\sigma$: $$\sigma = \frac{7}{0.84} \approx 8.33$$ 14. **Step:** Find $P(X < 330)$: Calculate $z = \frac{330 - 333}{8.33} = -0.36$. 15. **Step:** From tables, $P(Z < -0.36) \approx 0.36$, so about 36% of cans contain less than 330 ml. --- 16. **Problem:** $X \sim N(\mu, 12)$, $P(X > 32) = 0.8438$. 17. **Step:** Find $z$ such that $P(Z > z) = 0.8438$, so $P(Z < z) = 0.1562$. From tables, $z \approx -1.0$. 18. **Step:** Use $z = \frac{32 - \mu}{12} = -1.0$. 19. **Step:** Solve for $\mu$: $$32 - \mu = -12$$ $$\mu = 44$$ 20. **Step:** Find $P(34.5 < X < 35.3)$. 21. **Step:** Calculate $z$-scores: $$z_1 = \frac{34.5 - 44}{12} = -0.79$$ $$z_2 = \frac{35.3 - 44}{12} = -0.73$$ 22. **Step:** From tables, $P(Z < -0.73) = 0.2327$, $P(Z < -0.79) = 0.2148$. 23. **Step:** Probability between is $0.2327 - 0.2148 = 0.0179$ or 1.79%. --- 24. **Problem:** Heights of 500 people, $\sigma = 0.080$. 129 people have height between mean $\mu$ and 1.806 m. 25. **Step:** Proportion between $\mu$ and 1.806 is $129/500 = 0.258$. 26. **Step:** Since normal distribution is symmetric, $P(\mu < X < 1.806) = 0.258$. 27. **Step:** Find $z$ such that $P(0 < Z < z) = 0.258$. From tables, $z \approx 0.65$. 28. **Step:** Use $z = \frac{1.806 - \mu}{0.080} = 0.65$. 29. **Step:** Solve for $\mu$: $$1.806 - \mu = 0.65 \times 0.080 = 0.052$$ $$\mu = 1.806 - 0.052 = 1.754$$ --- 30. **Problem:** Masses of boxes, 20% heavier than 5.08 kg, 15% heavier than 5.62 kg. 31. **Step:** Let $X \sim N(\mu, \sigma)$. 32. **Step:** For 20% heavier than 5.08, $P(X > 5.08) = 0.20$, so $P(Z > z_1) = 0.20$, $z_1 = 0.84$. 33. **Step:** For 15% heavier than 5.62, $P(X > 5.62) = 0.15$, so $P(Z > z_2) = 0.15$, $z_2 = 1.04$. 34. **Step:** Write equations: $$\frac{5.08 - \mu}{\sigma} = 0.84$$ $$\frac{5.62 - \mu}{\sigma} = 1.04$$ 35. **Step:** Subtract equations: $$\frac{5.62 - 5.08}{\sigma} = 1.04 - 0.84 = 0.20$$ $$\frac{0.54}{\sigma} = 0.20$$ $$\sigma = \frac{0.54}{0.20} = 2.7$$ 36. **Step:** Use $\sigma$ to find $\mu$: $$0.84 = \frac{5.08 - \mu}{2.7}$$ $$5.08 - \mu = 0.84 \times 2.7 = 2.268$$ $$\mu = 5.08 - 2.268 = 2.812$$ --- 37. **Problem:** $X \sim N(\mu, \sigma)$, $P(X > 80) = 0.0113$, $P(X < 30) = 0.0287$. 38. **Step:** Find $z$ values: $$P(Z > z_1) = 0.0113 \Rightarrow z_1 = 2.26$$ $$P(Z < z_2) = 0.0287 \Rightarrow z_2 = -1.90$$ 39. **Step:** Write equations: $$\frac{80 - \mu}{\sigma} = 2.26$$ $$\frac{30 - \mu}{\sigma} = -1.90$$ 40. **Step:** Subtract equations: $$\frac{80 - 30}{\sigma} = 2.26 - (-1.90) = 4.16$$ $$\frac{50}{\sigma} = 4.16$$ $$\sigma = \frac{50}{4.16} \approx 12.02$$ 41. **Step:** Use $\sigma$ to find $\mu$: $$2.26 = \frac{80 - \mu}{12.02}$$ $$80 - \mu = 2.26 \times 12.02 = 27.17$$ $$\mu = 80 - 27.17 = 52.83$$ **Final answers:** 1. $\mu = 40$ 2. $\sigma \approx 10.71$ 3. (i) $\sigma \approx 8.33$, (ii) 36% 4. (i) $\mu = 44$, (ii) 1.79% 5. $\mu \approx 1.754$ 6. $\mu \approx 2.812$, $\sigma = 2.7$ 7. $\mu \approx 52.83$, $\sigma \approx 12.02$