\( X \) 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 \( 77 \) for a normally distributed variable with mean \( \mu = 41 \) and standard deviation \( \sigma = 22 \), we first need to convert \( 77 \) to a z-score using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Substituting the values: \[ z = \frac{(77 - 41)}{22} = \frac{36}{22} \approx 1.6364 \] Next, we use the z-score to find the probability that \( X < 77 \). We can look up \( z = 1.6364 \) in the standard normal distribution table or use a calculator. The cumulative probability corresponding to \( z = 1.6364 \) is approximately \( 0.9484 \). So, the probability that \( X \) is less than \( 77 \) is: \[ P(X < 77) \approx 0.948 \] Therefore, the final answer, rounded to the nearest thousandth, is \( \boxed{0.948} \).