\( X \) is a normally distributed random variable with mean 77 and standard deviation 9 . What is the probability that \( X \) is between 20 and 66 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 20 and 66 for a normally distributed random variable \( X \) with a mean (\( \mu \)) of 77 and a standard deviation (\( \sigma \)) of 9, follow these steps: 1. **Standardize the Values:** Convert the values 20 and 66 to their corresponding \( z \)-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 20 \): \[ z_1 = \frac{20 - 77}{9} = \frac{-57}{9} = -6.333 \] - For \( X = 66 \): \[ z_2 = \frac{66 - 77}{9} = \frac{-11}{9} \approx -1.222 \] 2. **Determine the Probability:** The probability \( P(20 \leq X \leq 66) \) is equivalent to \( P(z_1 \leq Z \leq z_2) \) in the standard normal distribution. - Since \( z_1 = -6.333 \) is far in the lower tail, \( \Phi(z_1) \) is practically 0. - Therefore, \( P(20 \leq X \leq 66) \approx \Phi(z_2) \). 3. **Look Up the Standard Normal CDF:** Using standard normal distribution tables or a calculator: \[ \Phi(-1.222) \approx 0.111 \] This means there's approximately an 11.1% chance that \( X \) falls between 20 and 66. **Final Answer:** The probability is approximately **0.111**.