\( X \) is a normally distributed random variable with mean 39 and standard deviation 12 . What is the probability that \( X \) is between 3 and 75 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To apply the \( 0.68-0.95-0.997 \) rule, we first need to determine how many standard deviations the values 3 and 75 are from the mean (39). 1. For \( X = 3 \): \[ Z = \frac{3 - 39}{12} = \frac{-36}{12} = -3 \] 2. For \( X = 75 \): \[ Z = \frac{75 - 39}{12} = \frac{36}{12} = 3 \] Now, we identify the probability that \( X \) falls between \( Z = -3 \) and \( Z = 3 \). According to the \( 0.68-0.95-0.997 \) rule: - Approximately \( 99.7\% \) of the data falls within \( 3 \) standard deviations from the mean. Therefore, the probability that \( X \) is between 3 and 75 is: \[ P(3 < X < 75) \approx 0.997 \] Rounded to the nearest thousandth, the probability is **0.997**.