QUESTION 1 Suppose that the number of movies people rent per month is normally distributed with a mean of 4.8 and a standard deviation of 1.5. What is the probability that in a random sample of 50 people that the average number of movies they rented last month is greater than 4.5 ? (please round your answer to 4 decimal places)

To determine the probability that the average number of movies rented by a random sample of 50 people is greater than 4.5, we'll use the properties of the normal distribution. **Given:** - Population mean (\( \mu \)) = 4.8 movies - Population standard deviation (\( \sigma \)) = 1.5 movies - Sample size (\( n \)) = 50 **Step 1: Calculate the Standard Error (SE)** \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{50}} \approx 0.2121 \] **Step 2: Compute the Z-score** \[ Z = \frac{\bar{X} - \mu}{SE} = \frac{4.5 - 4.8}{0.2121} \approx \frac{-0.3}{0.2121} \approx -1.4142 \] **Step 3: Find the Probability** We need to find \( P(\bar{X} > 4.5) \), which is equivalent to \( P(Z > -1.4142) \). Using standard normal distribution tables or a calculator: \[ P(Z > -1.4142) \approx 0.9214 \] **Conclusion:** The probability that the average number of movies rented is greater than 4.5 is **0.9214** (rounded to four decimal places). **Answer:** The probability is 0.9214