\( X \) is a normally distributed random variable with mean 94 and standard deviation 22. What is the probability that \( X \) is between 46 and 96 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that a normally distributed random variable \( X \) with mean \( \mu = 94 \) and standard deviation \( \sigma = 22 \) falls between 46 and 96, we'll follow these steps: 1. **Standardize the Values:** Convert the values 46 and 96 to their corresponding \( z \)-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 46 \): \[ z_1 = \frac{46 - 94}{22} = \frac{-48}{22} \approx -2.182 \] - For \( X = 96 \): \[ z_2 = \frac{96 - 94}{22} = \frac{2}{22} \approx 0.091 \] 2. **Find the Cumulative Probabilities:** Using standard normal distribution tables or a calculator: - \( \Phi(z_1) = \Phi(-2.182) \approx 0.0145 \) - \( \Phi(z_2) = \Phi(0.091) \approx 0.5363 \) Here, \( \Phi(z) \) represents the cumulative distribution function (CDF) for the standard normal distribution. 3. **Calculate the Probability Between 46 and 96:** \[ P(46 \leq X \leq 96) = \Phi(z_2) - \Phi(z_1) = 0.5363 - 0.0145 \approx 0.5218 \] 4. **Round to the Nearest Thousandth:** \[ P(46 \leq X \leq 96) \approx 0.522 \] **Final Answer:** \[ \boxed{0.522} \]