\( X \) is a normally distributed random variable with mean 89 and standard deviation 4 . What is the probability that \( X \) is between 87 and 91 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 87 and 91, where \( X \) is normally distributed with a mean (\( \mu \)) of 89 and a standard deviation (\( \sigma \)) of 4, follow these steps: 1. **Standardize the Values:** Convert the values 87 and 91 to their corresponding Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] - For \( X = 87 \): \[ Z_1 = \frac{87 - 89}{4} = -0.5 \] - For \( X = 91 \): \[ Z_2 = \frac{91 - 89}{4} = 0.5 \] 2. **Use the Standard Normal Distribution Table:** Find the probabilities corresponding to \( Z_1 = -0.5 \) and \( Z_2 = 0.5 \). - \( \Phi(0.5) \approx 0.6915 \) - \( \Phi(-0.5) = 1 - \Phi(0.5) \approx 1 - 0.6915 = 0.3085 \) 3. **Calculate the Probability Between the Two Z-scores:** \[ P(87 \leq X \leq 91) = \Phi(0.5) - \Phi(-0.5) \approx 0.6915 - 0.3085 = 0.383 \] Rounded to the nearest thousandth, the probability is **0.383**. **Answer:** 0.383