\( X \) 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 random variable \( X \) with a mean (\( \mu \)) of 41 and a standard deviation (\( \sigma \)) of 22, follow these steps: 1. **Standardize the Value:** Convert the value \( 77 \) to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{77 - 41}{22} = \frac{36}{22} \approx 1.636 \] 2. **Find the Cumulative Probability:** Using standard normal distribution tables or a calculator, determine \( \Phi(Z) \), which represents the probability that \( Z \) is less than 1.636. - For \( Z = 1.63 \), \( \Phi(1.63) \approx 0.9484 \). - For \( Z = 1.64 \), \( \Phi(1.64) \approx 0.9495 \). Since \( Z \approx 1.636 \) is closer to 1.64, we can interpolate: \[ \Phi(1.636) \approx 0.949 \] 3. **Conclusion:** The probability that \( X \) is less than \( 77 \) is approximately **0.949** when rounded to the nearest thousandth. **Final Answer:** The probability that \( X \) is less than \( 77 \) is approximately **0.949**.