\( X \) is a normally distributed random variable with mean 22 and standard deviation 9 . What is the probability that \( X \) is between 30 and \( 67 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that a normally distributed random variable \( X \) with mean \( \mu = 22 \) and standard deviation \( \sigma = 9 \) falls between 30 and 67, we can use the standard normal distribution (z-scores). **Step 1: Convert the values to z-scores.** The z-score formula is: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 30 \): \[ z_1 = \frac{30 - 22}{9} = \frac{8}{9} \approx 0.889 \] - For \( X = 67 \): \[ z_2 = \frac{67 - 22}{9} = \frac{45}{9} = 5 \] **Step 2: Find the probabilities corresponding to these z-scores.** Using standard normal distribution tables or a calculator: - \( P(Z < 0.889) \approx 0.8126 \) - \( P(Z < 5) \approx 1 \) (since a z-score of 5 is far in the tail, the probability is practically 1) **Step 3: Calculate the probability between the two z-scores.** \[ P(30 < X < 67) = P(Z < 5) - P(Z < 0.889) \approx 1 - 0.8126 = 0.1874 \] **Rounding to the nearest thousandth:** \[ P(30 < X < 67) \approx 0.187 \] **Answer:** The probability is approximately 0.187.