1. **Problem statement:** We have delivery times normally distributed with mean $\mu=5$ days and standard deviation $\sigma=1.2$ days. 2. **Formula and rules:** For a normal distribution, we use the standard normal variable $Z=\frac{X-\mu}{\sigma}$ to find probabilities from standard normal tables or a calculator. --- **a. Percentage of orders delivered in less than 4 days:** 3. Calculate $Z$ for $X=4$: $$Z=\frac{4-5}{1.2} = \frac{-1}{1.2} = -0.8333$$ 4. Find $P(X<4) = P(Z < -0.8333)$ using standard normal distribution. 5. From standard normal tables or calculator, $P(Z < -0.8333) \approx 0.2023$. 6. So, about 20.23% of orders are delivered in less than 4 days. --- **b. Proportion of orders delivered between 4 and 6 days:** 7. Calculate $Z$ for $X=6$: $$Z=\frac{6-5}{1.2} = \frac{1}{1.2} = 0.8333$$ 8. Find $P(4 < X < 6) = P(-0.8333 < Z < 0.8333)$. 9. Using symmetry and tables: $$P(Z < 0.8333) \approx 0.7977$$ $$P(Z < -0.8333) \approx 0.2023$$ 10. So, $$P(-0.8333 < Z < 0.8333) = 0.7977 - 0.2023 = 0.5954$$ 11. About 59.54% of orders take between 4 and 6 days. --- **c. Delivery time threshold for 95% of orders:** 12. We want $P(X \leq x) = 0.95$. 13. Find $Z$ such that $P(Z \leq z) = 0.95$. 14. From standard normal tables, $z_{0.95} \approx 1.645$. 15. Convert back to $X$: $$x = \mu + z \sigma = 5 + 1.645 \times 1.2 = 5 + 1.974 = 6.974$$ 16. So, the company should guarantee delivery within about 6.97 days to cover 95% of orders.