EE. 3 Find probabilities using the normal distribution I SSW \( X \) is a normally distributed random variable with mean 28 and standard deviation 20 . What is the probability that \( X \) is greater than 88 ? 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 greater than 88, we first need to calculate the z-score for \( X = 88 \). The z-score is given by the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu = 28 \) (mean) and \( \sigma = 20 \) (standard deviation). Plugging the values in: \[ z = \frac{(88 - 28)}{20} = \frac{60}{20} = 3 \] Next, we use the z-score to find the corresponding probability. The z-score of 3 indicates that 88 is 3 standard deviations above the mean. According to the 0.68-0.95-0.997 rule, approximately 99.7% of the data falls within 3 standard deviations of the mean, meaning that only about 0.3% is beyond this point. Thus, the probability that \( X \) is greater than 88 can be represented as: \[ P(X > 88) \approx 0.003 \] Rounding to the nearest thousandth: \[ \boxed{0.003} \]