\( X \) is a normally distributed random variable with mean 90 and standard deviation 21. What is the probability that \( X \) is between 9 and 50 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 9 and 50 for a normally distributed random variable with a mean of 90 and a standard deviation of 21, we can use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 9 \): \[ Z_1 = \frac{9 - 90}{21} = \frac{-81}{21} \approx -3.857 \] For \( X = 50 \): \[ Z_2 = \frac{50 - 90}{21} = \frac{-40}{21} \approx -1.905 \] Now, we look up these Z-scores in the standard normal distribution table or use a calculator to find the probabilities. For \( Z_1 \approx -3.857 \), the probability \( P(Z < -3.857) \) is approximately 0.00006 (very close to 0). For \( Z_2 \approx -1.905 \), the probability \( P(Z < -1.905) \) is approximately 0.0287. To find the probability that \( X \) is between 9 and 50, we calculate: \[ P(9 < X < 50) = P(Z < -1.905) - P(Z < -3.857) \approx 0.0287 - 0 = 0.0287 \] Thus, the probability that \( X \) is between 9 and 50 is approximately **0.029** when rounded to the nearest thousandth.