\( X \) is a normally distributed random variable with mean 36 and standard deviation 20 . What is the probability that \( X \) is between 16 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 16 and 96, we first need to consider the mean and standard deviation of the distribution. The mean (μ) is 36, and the standard deviation (σ) is 20. Now, we can determine how many standard deviations 16 and 96 are from the mean: 1. For 16: \[ z_1 = \frac{16 - 36}{20} = \frac{-20}{20} = -1 \] 2. For 96: \[ z_2 = \frac{96 - 36}{20} = \frac{60}{20} = 3 \] Using the \( 0.68-0.95-0.997 \) rule for normal distributions, we can assess probabilities based on these z-scores: - The probability that \( X \) is within 1 standard deviation of the mean (from \( \mu - \sigma \) to \( \mu + \sigma \), or 16 to 56) is approximately 68%. - The probability that \( X \) is within 3 standard deviations (from \( \mu - 3\sigma \) to \( \mu + 3\sigma \), or -24 to 96) is approximately 99.7%. Since we want the probability of \( X \) being between 16 and 96, we can combine these two ranges: 1. From 16 to 56 gives us 68%. 2. From 56 to 96 adds the area from 56 to 96, which is half of the remaining area outside of -1 to 3, about 31.7% (half of (99.7% - 68%)). Thus: \[ P(16 < X < 96) \approx 68\% + 31.7\% = 99.7\% \] Converting this to a decimal gives approximately \( 0.997 \). So, the final answer rounded to the nearest thousandth is: \[ \boxed{0.997} \]