\( X \) is a normally distributed random variable with mean 54 and standard deviation 3 . What is the probability that \( X \) is between 52 and 56 ? Write your answer as a decimal rounded to the nearest thousandth.

Given that \( X \) is normally distributed with a mean (\( \mu \)) of 54 and a standard deviation (\( \sigma \)) of 3, we want to find the probability that \( X \) lies between 52 and 56. To solve this, we'll use the **standard normal distribution** (Z-score): \[ Z = \frac{X - \mu}{\sigma} \] 1. **Calculate the Z-scores for 52 and 56:** - For \( X = 52 \): \[ Z_1 = \frac{52 - 54}{3} = -\frac{2}{3} \approx -0.6667 \] - For \( X = 56 \): \[ Z_2 = \frac{56 - 54}{3} = \frac{2}{3} \approx 0.6667 \] 2. **Find the cumulative probabilities for these Z-scores:** Using standard normal distribution tables or a calculator: - \( \Phi(0.6667) \approx 0.7475 \) - Since the normal distribution is symmetric, \( \Phi(-0.6667) = 1 - \Phi(0.6667) \approx 1 - 0.7475 = 0.2525 \) 3. **Calculate the probability between 52 and 56:** \[ P(52 \leq X \leq 56) = \Phi(0.6667) - \Phi(-0.6667) = 0.7475 - 0.2525 = 0.495 \] **Rounded to the nearest thousandth, the probability is 0.495.** **Answer:** 0.495