\( X \) is a normally distributed random variable with mean 80 and standard deviation 6 . What is the probability that \( X \) is between 70 and 93 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 70 and 93 for a normally distributed random variable \( X \) with mean \( \mu = 80 \) and standard deviation \( \sigma = 6 \), follow these steps: 1. **Standardize the Values:** Convert the values 70 and 93 to their corresponding z-scores using the formula: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 70 \): \[ z_1 = \frac{70 - 80}{6} = \frac{-10}{6} \approx -1.667 \] - For \( X = 93 \): \[ z_2 = \frac{93 - 80}{6} = \frac{13}{6} \approx 2.167 \] 2. **Find the Corresponding Probabilities:** Using standard normal distribution tables or a calculator: - \( P(Z \leq 2.167) \approx 0.9846 \) - \( P(Z \leq -1.667) \approx 0.0485 \) 3. **Calculate the Probability Between 70 and 93:** \[ P(70 \leq X \leq 93) = P(Z \leq 2.167) - P(Z \leq -1.667) \approx 0.9846 - 0.0485 = 0.9361 \] 4. **Round to the Nearest Thousandth:** \[ P(70 \leq X \leq 93) \approx 0.936 \] **Final Answer:** The probability is approximately **0.936**.