\( X \) is a normally distributed random variable with mean 27 and standard deviation 13 . What is the probability that \( X \) is between 19 and \( 69 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that a normally distributed random variable \( X \) is between 19 and 69, we can use the z-score formula and the standard normal distribution table. Given: - Mean (\( \mu \)) = 27 - Standard Deviation (\( \sigma \)) = 13 We need to find the probability that \( X \) is between 19 and 69. First, let's calculate the z-scores for 19 and 69: \[ z = \frac{X - \mu}{\sigma} \] For \( X = 19 \): \[ z = \frac{19 - 27}{13} = \frac{-8}{13} \] For \( X = 69 \): \[ z = \frac{69 - 27}{13} = \frac{42}{13} \] Now, we can use the standard normal distribution table to find the probabilities corresponding to these z-scores. Let's calculate the probabilities: - Probability that \( X \) is less than 19: \[ P(X < 19) = P(Z < -8/13) \] - Probability that \( X \) is less than 69: \[ P(X < 69) = P(Z < 42/13) \] Using the standard normal distribution table, we can find the probabilities: - \( P(Z < -8/13) \approx 0.5000 \) - \( P(Z < 42/13) \approx 0.9999 \) Now, we can find the probability that \( X \) is between 19 and 69: \[ P(19 < X < 69) = P(X < 69) - P(X < 19) \] Substitute the values: \[ P(19 < X < 69) = 0.9999 - 0.5000 = 0.4999 \] Therefore, the probability that \( X \) is between 19 and 69 is approximately 0.4999, which rounded to the nearest thousandth is 0.500.