\( X \) 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 convert the value to a z-score using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Plugging in the values: \[ z = \frac{(77 - 41)}{22} = \frac{36}{22} \approx 1.6364 \] Next, we can use a standard normal distribution table or a calculator to find the probability corresponding to the z-score of approximately \( 1.6364 \). Using the z-table or calculator: \[ P(Z < 1.6364) \approx 0.9484 \] Thus, the probability that \( X \) is less than \( 77 \) is approximately \( 0.948 \) when rounded to the nearest thousandth.