\( X \) is a normally distributed random variable with mean 39 and standard deviation 11 . What is the probability that \( X \) is greater than 17 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To determine the probability that \( X \) is greater than 17, where \( X \) is normally distributed with a mean (\( \mu \)) of 39 and a standard deviation (\( \sigma \)) of 11, we can use the \( 0.68-0.95-0.997 \) empirical rule. 1. **Calculate the Z-score:** \[ Z = \frac{X - \mu}{\sigma} = \frac{17 - 39}{11} = \frac{-22}{11} = -2 \] 2. **Interpret the Z-score using the empirical rule:** - The empirical rule states that approximately 95% of the data lies within \( \mu \pm 2\sigma \). - Therefore, \( P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) \approx 0.95 \). 3. **Determine the probability \( P(X > 17) \):** - \( \mu - 2\sigma = 39 - 2 \times 11 = 17 \). - The probability that \( X \) is less than 17 is approximately \( 2.5\% \) (since 5% is split equally in the two tails of the distribution). - Hence, the probability that \( X \) is greater than 17 is: \[ P(X > 17) = 1 - P(X < 17) = 1 - 0.025 = 0.975 \] **Final Answer:** The probability that \( X \) is greater than 17 is **0.975**.