\( X \) 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 a normally distributed random variable \( X \) is less than a specific value, we can use the z-score formula and a standard normal distribution table (also known as a z-table). Given: - Mean (\( \mu \)) = 41 - Standard Deviation (\( \sigma \)) = 22 - Value of interest (\( x \)) = 77 First, we need to calculate the z-score for \( x = 77 \) using the formula: \[ z = \frac{x - \mu}{\sigma} \] Substitute the given values into the formula: \[ z = \frac{77 - 41}{22} \] Now, we can calculate the z-score: \[ z = \frac{36}{22} \] \[ z = 1.6363 \] Next, we use the z-score to find the probability that \( X \) is less than 77. We look up the z-score in a standard normal distribution table to find the corresponding probability. The probability that \( X \) is less than 77 is the same as the probability that \( z \) is less than 1.6363. Using a standard normal distribution table, we find that the probability is approximately 0.948. Therefore, the probability that \( X \) is less than 77 is approximately 0.948, rounded to the nearest thousandth.