\( X \) is a normally distributed random variable with mean 77 and standard deviation 9 . What is the probability that \( X \) is between 20 and 66 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 20 and 66 for a normally distributed random variable, we first standardize these values using the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \( \mu = 77 \) (mean) and \( \sigma = 9 \) (standard deviation). First, we calculate the Z-scores for 20 and 66: For \( X = 20 \): \[ Z_{20} = \frac{(20 - 77)}{9} = \frac{-57}{9} \approx -6.33 \] For \( X = 66 \): \[ Z_{66} = \frac{(66 - 77)}{9} = \frac{-11}{9} \approx -1.22 \] Now, we use the Z-table (or a calculator) to find the probabilities associated with these Z-scores. The probability corresponding to \( Z = -6.33 \) is virtually 0 since it's extremely far in the left tail of the normal distribution. The probability corresponding to \( Z = -1.22 \) is approximately 0.1103. Now we find the probability that \( X \) is between 20 and 66: \[ P(20 < X < 66) = P(Z < -1.22) - P(Z < -6.33) \approx 0.1103 - 0 \approx 0.1103 \] Thus, the probability that \( X \) is between 20 and 66 is approximately \( \boxed{0.110} \) when rounded to the nearest thousandth.