\( X \) is a normally distributed random variable with mean 18 and standard deviation 14 . What is the probability that \( X \) is greater than 43 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is greater than 43, we can first standardize the variable using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Where: - \( \mu = 18 \) (mean) - \( \sigma = 14 \) (standard deviation) - \( X = 43 \) Calculating the Z-score: \[ Z = \frac{43 - 18}{14} = \frac{25}{14} \approx 1.786 \] Next, we look up the Z-score in a standard normal distribution table or use a calculator to find the probability that \( Z \) is less than 1.786. The corresponding value is approximately 0.962. To find the probability that \( X \) is greater than 43, we subtract this value from 1: \[ P(X > 43) = 1 - P(Z < 1.786) \approx 1 - 0.962 = 0.038 \] So, the probability that \( X \) is greater than 43 is approximately 0.038 when rounded to the nearest thousandth.