\( X \) is a normally distributed random variable with mean 41 and standard deviation 22 . What is the probability that \( X \) is less than \( 77 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is less than \( 77 \), we first need to standardize the variable using the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] Here, \( \mu = 41 \) and \( \sigma = 22 \). Thus, we substitute in the values for \( X = 77 \): \[ Z = \frac{(77 - 41)}{22} = \frac{36}{22} \approx 1.636 \] Now, we can look up the Z-score of \( 1.636 \) in the standard normal distribution table, or use a calculator or software to find the cumulative probability. The cumulative probability for \( Z \approx 1.636 \) is approximately \( 0.9484 \). Therefore, the probability that \( X \) is less than \( 77 \) is: \[ P(X < 77) \approx 0.948 \] So, rounding to the nearest thousandth, the answer is \( 0.948 \).