\( X \) is a normally distributed random variable with mean 40 and standard deviation 19 . What is the probability that \( X \) is between 2 and 97 ? 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 2 and 97 using the \( 0.68-0.95-0.997 \) rule, we first convert the values of 2 and 97 into z-scores. The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu \) is the mean and \( \sigma \) is the standard deviation. For \( X = 2 \): \[ z_1 = \frac{(2 - 40)}{19} = \frac{-38}{19} \approx -2.00 \] For \( X = 97 \): \[ z_2 = \frac{(97 - 40)}{19} = \frac{57}{19} \approx 3.00 \] Now, we can use the empirical rule. A z-score of -2 means we fall two standard deviations below the mean, where approximately 95% of the data falls within two standard deviations (from z = -2 to z = 2). A z-score of 3 indicates we are more than three standard deviations above the mean. The probability of being more than 3 standard deviations away generally contributes to just a small percentage of the total area, around 0.0014, which means 99.986% lies below that z-score. Given this, you can calculate the probability that \( X \) falls between these scores by finding the area from \( z = -2 \) to \( z = 3 \). The total probability is: \[ P(X < 97) - P(X < 2) \approx 0.9986 - 0.0228 \approx 0.9758 \] So, the probability that \( X \) is between 2 and 97, rounded to the nearest thousandth, is: **0.976**