\( X \) is a normally distributed random variable with mean 65 and standard deviation 20 . What is the probability that \( X \) is greater than 10 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is greater than 10 for a normally distributed random variable with mean \( \mu = 65 \) and standard deviation \( \sigma = 20 \), we first need to convert the value 10 into a standard score (z-score). The z-score is calculated using the formula: \[ z = \frac{X - \mu}{\sigma} \] Substituting the known values: \[ z = \frac{10 - 65}{20} = \frac{-55}{20} = -2.75 \] Next, we need to check the z-table (standard normal distribution table) for the probability corresponding to \( z = -2.75 \). This gives us the probability that \( X \) is less than 10. Looking this up, we find: \[ P(Z < -2.75) \approx 0.003 \] To find the probability that \( X \) is greater than 10, we use: \[ P(X > 10) = 1 - P(X < 10) = 1 - P(Z < -2.75) \] Calculating that gives: \[ P(X > 10) = 1 - 0.003 = 0.997 \] Thus, the probability that \( X \) is greater than 10, rounded to the nearest thousandth, is: \[ \boxed{0.997} \]