\( X \) is a normally distributed random variable with mean 96 and standard deviation 22 . What is the probability that \( X \) is between 30 and 52 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To determine the probability that \( X \) is between 30 and 52, we first need to understand how these values relate to the mean and standard deviation. Given the mean (\( \mu \)) is 96 and the standard deviation (\( \sigma \)) is 22, we can calculate the Z-scores for both values. The Z-score formula is: \[ Z = \frac{(X - \mu)}{\sigma} \] Calculating the Z-scores: For \( 30 \): \[ Z_{30} = \frac{(30 - 96)}{22} = \frac{-66}{22} = -3 \] For \( 52 \): \[ Z_{52} = \frac{(52 - 96)}{22} = \frac{-44}{22} = -2 \] Now we need to find the probability corresponding to these Z-scores. Using the empirical rule (also known as the 68-95-99.7 rule), we know that: - Approximately 99.7% of the data lies within 3 standard deviations from the mean. - Approximately 95% lies within 2 standard deviations. Since both Z-scores (-2 and -3) fall outside the 1 standard deviation range but less than 3, we can analyze the intervals covered. 1. The range of \( X \) values between \( \mu - 2\sigma \) and \( \mu + 2\sigma \) (around 72 to 120) covers approximately 95% of the distribution. 2. The range of \( X \) values from \( \mu - 3\sigma \) to \( \mu + 3\sigma \) covers approximately 99.7%, but we are primarily interested in the area to the left of \( 52 \). To estimate the exact probability that \( X \) lies between 30 and 52: - The probability of being below a Z-score of -3 (30) is extremely small (close to 0). - The probability of being between Z-scores -3 and -2 (30 to 52) is between the probabilities corresponding to those Z-scores. The rough estimation of the probability will be very low since: - Beyond Z = -3, the cumulative probability is nearly 0. - Only a small area (the tail of the distribution) exists between these two Z-scores. Summarily, the cumulative probability from -3 to -2 contributes very little to the total, leading to a small probability. Thus, rounding to the nearest thousandth: \[ \text{Probability}(30 < X < 52) \approx 0.000 (or very minimally affecting the total). \] Therefore, we conclude that the probable decimal probability of \( X \) being between 30 and 52 is effectively 0.000.