\( X \) is a normally distributed random variable with mean 33 and standard deviation 9. What is the probability that \( X \) is less than 55 ? Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that \( X \) is less than 55 for a normally distributed random variable \( X \) with a mean (\( \mu \)) of 33 and a standard deviation (\( \sigma \)) of 9, follow these steps: 1. **Standardize the Value:** Convert the value 55 to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} \] \[ Z = \frac{55 - 33}{9} = \frac{22}{9} \approx 2.444 \] 2. **Find the Cumulative Probability:** Using standard normal distribution tables or a calculator, find the probability \( P(Z < 2.444) \). - For \( Z = 2.44 \), \( \Phi(2.44) \approx 0.9926 \) - For \( Z = 2.45 \), \( \Phi(2.45) \approx 0.9929 \) Since 2.444 is very close to 2.44, we can interpolate: \[ P(Z < 2.444) \approx 0.9926 + (0.444 \times 0.0003) \approx 0.9927 \] 3. **Round the Result:** Rounding 0.9927 to the nearest thousandth gives: \[ 0.993 \] **Final Answer:** The probability that \( X \) is less than 55 is **0.993**.