\( X \) is a normally distributed random variable with mean 4 and standard deviation 2 . What is the probability that \( X \) is between 9 and 14 ? Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that \( X \) is between 9 and 14, where \( X \) is normally distributed with a mean (\( \mu \)) of 4 and a standard deviation (\( \sigma \)) of 2, follow these steps: 1. **Standardize the Values:** Convert the values 9 and 14 to their corresponding z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 9 \): \[ z_1 = \frac{9 - 4}{2} = \frac{5}{2} = 2.5 \] - For \( X = 14 \): \[ z_2 = \frac{14 - 4}{2} = \frac{10}{2} = 5.0 \] 2. **Find the Probabilities Using the Standard Normal Distribution:** - The probability that \( Z \) is less than 2.5 (\( \Phi(2.5) \)) is approximately 0.9938. - The probability that \( Z \) is less than 5.0 (\( \Phi(5.0) \)) is essentially 1 since a z-score of 5 is extremely high. 3. **Calculate the Desired Probability:** \[ P(9 \leq X \leq 14) = P(2.5 \leq Z \leq 5) = \Phi(5) - \Phi(2.5) \approx 1 - 0.9938 = 0.0062 \] Rounding to the nearest thousandth: \[ 0.0062 \approx 0.006 \] **Final Answer:** \(\boxed{0.006}\)