\( X \) is a normally distributed random variable with mean 42 and standard deviation 18 . What is the probability that \( X \) is between 6 and 96 ? 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 6 and 96, we can use the 68-95-99.7 rule for a normal distribution. Given: - Mean (\( \mu \)) = 42 - Standard Deviation (\( \sigma \)) = 18 The 68-95-99.7 rule states that: - About 68% of the data falls within 1 standard deviation of the mean. - About 95% of the data falls within 2 standard deviations of the mean. - About 99.7% of the data falls within 3 standard deviations of the mean. Since we are interested in the probability that \( X \) is between 6 and 96, we need to find the z-scores for these values. The z-score formula is: \[ z = \frac{X - \mu}{\sigma} \] Let's calculate the z-scores for 6 and 96: - For \( X = 6 \): \[ z = \frac{6 - 42}{18} = \frac{-36}{18} = -2 \] - For \( X = 96 \): \[ z = \frac{96 - 42}{18} = \frac{54}{18} = 3 \] Now, we can use the z-scores to find the probabilities using a standard normal distribution table or calculator. The probability that \( X \) is between 6 and 96 is the area under the standard normal curve between z = -2 and z = 3. Using a standard normal distribution table or calculator, we find that the area between z = -2 and z = 3 is approximately 0.997. Therefore, the probability that \( X \) is between 6 and 96 is approximately 0.997.