\( X \) 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, where \( X \) is normally distributed with a mean (\( \mu \)) of 90 and a standard deviation (\( \sigma \)) of 21, we can follow these steps: 1. **Standardize the Values:** Convert the values 9 and 50 to their corresponding \( z \)-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 9 \): \[ z_1 = \frac{9 - 90}{21} = \frac{-81}{21} \approx -3.857 \] - For \( X = 50 \): \[ z_2 = \frac{50 - 90}{21} = \frac{-40}{21} \approx -1.905 \] 2. **Find the Corresponding Probabilities:** Using standard normal distribution tables or a calculator: - \( \Phi(z_2) \) for \( z_2 \approx -1.905 \) is approximately 0.0286. - \( \Phi(z_1) \) for \( z_1 \approx -3.857 \) is very close to 0 (specifically, about 0.000056). 3. **Calculate the Probability Between the Two \( z \)-scores:** \[ P(9 \leq X \leq 50) = \Phi(z_2) - \Phi(z_1) \approx 0.0286 - 0.000056 \approx 0.0285 \] 4. **Round to the Nearest Thousandth:** \[ 0.0285 \text{ rounded to the nearest thousandth is } 0.029 \] **Final Answer:** The probability is **0.029**.