\( X \) is a normally distributed random variable with mean 9 and standard deviation 23. What is the probability that \( X \) is between 21 and 91 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 21 and 91, where \( X \) is normally distributed with a mean (\( \mu \)) of 9 and a standard deviation (\( \sigma \)) of 23, we can follow these steps: 1. **Standardize the Values:** Convert the values 21 and 91 to their corresponding \( Z \)-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] - For \( X = 21 \): \[ Z_1 = \frac{21 - 9}{23} = \frac{12}{23} \approx 0.522 \] - For \( X = 91 \): \[ Z_2 = \frac{91 - 9}{23} = \frac{82}{23} \approx 3.565 \] 2. **Find the Probabilities:** Using standard normal distribution tables or a calculator: - \( \Phi(Z_1) = \Phi(0.522) \approx 0.699 \) - \( \Phi(Z_2) = \Phi(3.565) \approx 0.9998 \) 3. **Calculate the Probability Between 21 and 91:** \[ P(21 \leq X \leq 91) = \Phi(Z_2) - \Phi(Z_1) \approx 0.9998 - 0.699 = 0.3008 \] Rounding to the nearest thousandth: \[ P(21 \leq X \leq 91) \approx 0.301 \] **Answer:** The probability that \( X \) is between 21 and 91 is approximately **0.301**.