\( X \) is a normally distributed random variable with mean 26 and standard deviation 9 . What is the probability that \( X \) is greater than \( 6 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that \( X \) is greater than 6, where \( X \) is normally distributed with a mean (\( \mu \)) of 26 and a standard deviation (\( \sigma \)) of 9, follow these steps: 1. **Standardize the Variable:** Convert \( X \) to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 6 \): \[ Z = \frac{6 - 26}{9} = \frac{-20}{9} \approx -2.222 \] 2. **Find the Probability:** We need \( P(X > 6) \), which is equivalent to \( P(Z > -2.222) \). Using the symmetry of the normal distribution: \[ P(Z > -2.222) = P(Z < 2.222) \] 3. **Lookup the Standard Normal CDF:** Using standard normal distribution tables or a calculator: \[ \Phi(2.222) \approx 0.9865 \] 4. **Round to the Nearest Thousandth:** \[ 0.9865 \text{ rounded to the nearest thousandth is } 0.987 \] **Final Answer:** The probability that \( X \) is greater than 6 is **0.987**.