1. **Problem statement:** We have a normal distribution for cosmic radiation exposure with mean $\mu = 4.35$ mrem and standard deviation $\sigma = 0.59$ mrem. We want to find probabilities for two cases: (i) The exposure is between 4.00 and 5.00 mrem. (ii) The exposure is at least 5.50 mrem. 2. **Formula and rules:** For a normal random variable $X \sim N(\mu, \sigma^2)$, the standardized variable is $$Z = \frac{X - \mu}{\sigma}$$ which follows the standard normal distribution $N(0,1)$. We use the standard normal table or a calculator to find probabilities. 3. **Calculate for (i):** Convert 4.00 and 5.00 to $Z$-scores: $$Z_1 = \frac{4.00 - 4.35}{0.59} = \frac{-0.35}{0.59} \approx -0.5932$$ $$Z_2 = \frac{5.00 - 4.35}{0.59} = \frac{0.65}{0.59} \approx 1.1017$$ Find probabilities from standard normal table: $$P(4.00 < X < 5.00) = P(Z_1 < Z < Z_2) = P(Z < 1.1017) - P(Z < -0.5932)$$ Using standard normal CDF values: $$P(Z < 1.1017) \approx 0.8643$$ $$P(Z < -0.5932) = 1 - P(Z < 0.5932) \approx 1 - 0.7236 = 0.2764$$ So, $$P(4.00 < X < 5.00) = 0.8643 - 0.2764 = 0.5879$$ 4. **Calculate for (ii):** Convert 5.50 to $Z$-score: $$Z = \frac{5.50 - 4.35}{0.59} = \frac{1.15}{0.59} \approx 1.9492$$ Find probability: $$P(X \geq 5.50) = P(Z \geq 1.9492) = 1 - P(Z < 1.9492)$$ From standard normal table: $$P(Z < 1.9492) \approx 0.9744$$ Therefore, $$P(X \geq 5.50) = 1 - 0.9744 = 0.0256$$ **Final answers:** (i) Probability exposure is between 4.00 and 5.00 mrem is approximately $0.588$. (ii) Probability exposure is at least 5.50 mrem is approximately $0.026$. These probabilities tell us how likely it is for a person to receive radiation in those ranges during a flight.