\( X \) 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 (the empirical rule), we first need to standardize the values of 2 and 97 into Z-scores. The formula for Z-score is: \[ Z = \frac{(X - \mu)}{\sigma} \] Where \( \mu \) is the mean and \( \sigma \) is the standard deviation. 1. For \( X = 2 \): \[ Z = \frac{(2 - 40)}{19} = \frac{-38}{19} \approx -2 \] 2. For \( X = 97 \): \[ Z = \frac{(97 - 40)}{19} = \frac{57}{19} \approx 3 \] Now, we can examine the Z-scores: - A Z-score of -2 corresponds to the lower end of the distribution, and according to the empirical rule, approximately 95% of the data lies within two standard deviations from the mean. - A Z-score of 3 falls well above the mean, where the area to the left of Z=3 is approximately 0.9987, meaning about 99.87% of the distribution lies below this point. To find the probability of \( X \) being between 2 and 97, we calculate: \[ P(2 < X < 97) = P(X < 97) - P(X < 2) \] \[ = 0.9987 - (1 - 0.9545) \] \[ = 0.9987 - 0.0455 \approx 0.9532 \] Rounding to the nearest thousandth gives us \( \approx 0.953 \). Therefore, the probability that \( X \) is between 2 and 97 is approximately 0.953.