7.4 Suppose the average length of stay in a chronic disease hospital of a certain type of patient is 60 days with a standard deviation of 15 . If it is reasonable to assume an approximately normal distribution of lengths of stay, find the probability that a randomly selected patient from this group will have a length of stay: \( \begin{array}{ll}\text { (a) Greater than } 50 \text { days } & \text { (b) Less than } 30 \text { days } \\ \text { (c) Between } 30 \text { and } 60 \text { days } & \text { (d) Greater than } 90 \text { days }\end{array} \)

To solve these probability questions, we'll use the properties of the **normal distribution**. Given: - **Mean (μ)** = 60 days - **Standard Deviation (σ)** = 15 days We'll calculate the **z-score** for each scenario and then use the standard normal distribution table (z-table) to find the corresponding probabilities. ### (a) Probability that a patient stays **greater than 50 days** 1. **Calculate the z-score:** \[ z = \frac{X - \mu}{\sigma} = \frac{50 - 60}{15} = \frac{-10}{15} = -0.6667 \] 2. **Find the probability:** \[ P(X > 50) = P(Z > -0.6667) \] Using the z-table, \[ P(Z < 0.6667) \approx 0.7475 \] Since the normal distribution is symmetric, \[ P(Z > -0.6667) = P(Z < 0.6667) \approx 74.75\% \] **Answer:** Approximately **74.8%** probability. --- ### (b) Probability that a patient stays **less than 30 days** 1. **Calculate the z-score:** \[ z = \frac{30 - 60}{15} = \frac{-30}{15} = -2 \] 2. **Find the probability:** \[ P(X < 30) = P(Z < -2) \approx 0.0228 \] **Answer:** Approximately **2.28%** probability. --- ### (c) Probability that a patient stays **between 30 and 60 days** 1. **Calculate the z-scores:** - For 30 days: \[ z_1 = \frac{30 - 60}{15} = -2 \] - For 60 days: \[ z_2 = \frac{60 - 60}{15} = 0 \] 2. **Find the probability:** \[ P(30 < X < 60) = P(-2 < Z < 0) \] Using the z-table, \[ P(Z < 0) = 0.5 \quad \text{and} \quad P(Z < -2) \approx 0.0228 \] \[ P(-2 < Z < 0) = 0.5 - 0.0228 = 0.4772 \] **Answer:** Approximately **47.7%** probability. --- ### (d) Probability that a patient stays **greater than 90 days** 1. **Calculate the z-score:** \[ z = \frac{90 - 60}{15} = \frac{30}{15} = 2 \] 2. **Find the probability:** \[ P(X > 90) = P(Z > 2) \approx 1 - 0.9772 = 0.0228 \] **Answer:** Approximately **2.28%** probability. --- ### **Summary of Probabilities:** - **(a)** \( P(X > 50) \approx 74.8\% \) - **(b)** \( P(X < 30) \approx 2.28\% \) - **(c)** \( P(30 < X < 60) \approx 47.7\% \) - **(d)** \( P(X > 90) \approx 2.28\% \)