\( X \) is a normally distributed random variable with mean 76 and standard deviation 4 . What is the probability that \( X \) is between 74 and 81 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 74 and 81, where \( X \) is normally distributed with a mean (\( \mu \)) of 76 and a standard deviation (\( \sigma \)) of 4, we can follow these steps: 1. **Standardize the Values:** Convert the values 74 and 81 to their corresponding z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 74 \): \[ z_1 = \frac{74 - 76}{4} = \frac{-2}{4} = -0.5 \] - For \( X = 81 \): \[ z_2 = \frac{81 - 76}{4} = \frac{5}{4} = 1.25 \] 2. **Find the Cumulative Probabilities:** Using standard normal distribution tables or a calculator: - \( \Phi(z_1) = \Phi(-0.5) \approx 0.3085 \) - \( \Phi(z_2) = \Phi(1.25) \approx 0.8944 \) 3. **Calculate the Probability Between the Two Z-scores:** \[ P(74 \leq X \leq 81) = \Phi(z_2) - \Phi(z_1) = 0.8944 - 0.3085 = 0.5859 \] 4. **Round to the Nearest Thousandth:** \[ P(74 \leq X \leq 81) \approx 0.586 \] **Final Answer:** The probability is 0.586