\( X \) 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 determine the probability that \( X \) is between 47 and 81 for a normally distributed random variable \( X \) with mean \( \mu = 98 \) and standard deviation \( \sigma = 17 \), we'll use the \( 0.68-0.95-0.997 \) empirical rule. ### Step 1: Calculate Z-scores First, convert the values 47 and 81 to their corresponding Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] - For \( X = 47 \): \[ Z_1 = \frac{47 - 98}{17} = \frac{-51}{17} = -3 \] - For \( X = 81 \): \[ Z_2 = \frac{81 - 98}{17} = \frac{-17}{17} = -1 \] ### Step 2: Determine Probabilities Using the Empirical Rule The empirical rule tells us: - Approximately **68%** of the data lies within **±1σ** of the mean. - Approximately **95%** within **±2σ**. - Approximately **99.7%** within **±3σ**. However, we are interested in the probability between **-3σ** and **-1σ**. Using standard normal distribution tables or properties: - \( P(Z \leq -1) \approx 0.1587 \) - \( P(Z \leq -3) \approx 0.00135 \) Thus, the probability between \( Z = -3 \) and \( Z = -1 \) is: \[ P(-3 \leq Z \leq -1) = P(Z \leq -1) - P(Z \leq -3) \approx 0.1587 - 0.00135 = 0.15735 \] Rounding to the nearest thousandth: \[ P(47 \leq X \leq 81) \approx 0.157 \] **Answer:** The probability is approximately 0.157.