\( X \) 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, we can first standardize these values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu = 76 \) and \( \sigma = 4 \). For \( X = 74 \): \[ z_{74} = \frac{(74 - 76)}{4} = \frac{-2}{4} = -0.5 \] For \( X = 81 \): \[ z_{81} = \frac{(81 - 76)}{4} = \frac{5}{4} = 1.25 \] Next, we'll look up these z-scores in the standard normal distribution table or use a calculator for the cumulative probabilities: - \( P(Z < -0.5) \approx 0.3085 \) - \( P(Z < 1.25) \approx 0.8944 \) Now we can find the probability that \( X \) is between 74 and 81: \[ P(74 < X < 81) = P(Z < 1.25) - P(Z < -0.5) \approx 0.8944 - 0.3085 = 0.5859 \] So, the probability that \( X \) is between 74 and 81 is approximately \( 0.586 \) when rounded to the nearest thousandth.