\( X \) is a normally distributed random variable with mean 23 and standard deviation 2 . What is the probability that \( X \) is between 24 and 96 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 24 and 96 for a normally distributed random variable \( X \) with mean \( \mu = 23 \) and standard deviation \( \sigma = 2 \), follow these steps: 1. **Standardize the Values:** Convert the values 24 and 96 to their corresponding z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 24 \): \[ z_1 = \frac{24 - 23}{2} = 0.5 \] - For \( X = 96 \): \[ z_2 = \frac{96 - 23}{2} = 36.5 \] 2. **Find the Probability:** - The z-score of 36.5 is extremely high, meaning \( P(X \leq 96) \) is virtually 1. - The probability that \( X \) is less than 24 corresponds to the cumulative distribution function (CDF) of the standard normal distribution at \( z = 0.5 \), which is approximately 0.6915. 3. **Calculate the Desired Probability:** \[ P(24 \leq X \leq 96) = P(X \leq 96) - P(X < 24) \approx 1 - 0.6915 = 0.3085 \] Rounding to the nearest thousandth: \[ 0.3085 \approx 0.309 \] **Answer:** 0.309