\( X \) is a normally distributed random variable with mean 98 and standard deviation 17 . What is the probability that \( X \) is between 47 and 81 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To find the probability that \( X \) is between 47 and 81 for a normally distributed random variable with mean \( \mu = 98 \) and standard deviation \( \sigma = 17 \), we can use the properties of the normal distribution. 1. **Calculate the Z-scores** for the values 47 and 81 using the formula: \[ Z = \frac{X - \mu}{\sigma} \] 2. **Calculate the Z-score for 47**: \[ Z_{47} = \frac{47 - 98}{17} \] 3. **Calculate the Z-score for 81**: \[ Z_{81} = \frac{81 - 98}{17} \] 4. **Use the Z-scores to find the probabilities** associated with these Z-scores using the standard normal distribution. Let's perform these calculations step by step. ### Step 1: Calculate Z-scores Calculating \( Z_{47} \): \[ Z_{47} = \frac{47 - 98}{17} = \frac{-51}{17} \approx -3 \] Calculating \( Z_{81} \): \[ Z_{81} = \frac{81 - 98}{17} = \frac{-17}{17} = -1 \] ### Step 2: Find the probabilities Now we need to find the probabilities corresponding to these Z-scores. - The probability of \( Z < -3 \) is very small (approximately 0.0013). - The probability of \( Z < -1 \) is approximately 0.1587. ### Step 3: Calculate the probability that \( X \) is between 47 and 81 To find the probability that \( X \) is between 47 and 81, we subtract the probability of \( Z < -3 \) from the probability of \( Z < -1 \): \[ P(47 < X < 81) = P(Z < -1) - P(Z < -3) \approx 0.1587 - 0.0013 \] Calculating this gives: \[ P(47 < X < 81) \approx 0.1574 \] ### Final Answer Thus, the probability that \( X \) is between 47 and 81 is approximately \( 0.157 \) when rounded to the nearest thousandth. \[ \boxed{0.157} \]