\( X \) is a normally distributed random variable with mean 78 and standard deviation 21 . What is the probability that \( X \) is between 73 and 83 ? Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that the normally distributed random variable \( X \) with mean \( \mu = 78 \) and standard deviation \( \sigma = 21 \) falls between 73 and 83, we can follow these steps: 1. **Standardize the Values:** Convert the values 73 and 83 to their corresponding z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 73 \): \[ z_1 = \frac{73 - 78}{21} = \frac{-5}{21} \approx -0.238 \] - For \( X = 83 \): \[ z_2 = \frac{83 - 78}{21} = \frac{5}{21} \approx 0.238 \] 2. **Find the Cumulative Probability:** Using standard normal distribution tables or a calculator, find the cumulative probability for \( z = 0.238 \). - \( \Phi(0.238) \approx 0.594 \) 3. **Calculate the Desired Probability:** The probability that \( X \) is between 73 and 83 is the difference between the cumulative probabilities at \( z_2 \) and \( z_1 \): \[ P(73 \leq X \leq 83) = \Phi(z_2) - \Phi(z_1) = 0.594 - (1 - 0.594) = 2 \times 0.594 - 1 = 0.188 \] **Final Answer:** \[ \boxed{0.188} \]