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 determine the probability that \( X \) is greater than 88 for a normally distributed random variable \( X \) with mean \( \mu = 28 \) and standard deviation \( \sigma = 20 \), we can use the **68-95-99.7 rule** (also known as the empirical rule). This rule provides a quick way to estimate probabilities for normal distributions. ### Steps: 1. **Calculate the Z-score:** The Z-score tells us how many standard deviations away 88 is from the mean. \[ Z = \frac{X - \mu}{\sigma} = \frac{88 - 28}{20} = \frac{60}{20} = 3 \] So, 88 is **3 standard deviations** above the mean. 2. **Apply the 68-95-99.7 Rule:** - **Within 1 standard deviation** (μ ± σ): ~68% - **Within 2 standard deviations** (μ ± 2σ): ~95% - **Within 3 standard deviations** (μ ± 3σ): ~99.7% This means that approximately **99.7%** of the data lies within 3 standard deviations of the mean. Therefore, the probability of \( X \) being **greater than** 3 standard deviations above the mean (i.e., \( X > 88 \)) is the remaining percentage in the upper tail. 3. **Calculate the Upper Tail Probability:** \[ P(X > 88) = \frac{1 - 0.997}{2} = \frac{0.003}{2} = 0.0015 \] Since probabilities must be expressed as decimals and rounded to the nearest thousandth: \[ 0.0015 \approx 0.002 \] ### Final Answer: The probability that \( X \) is greater than 88 is **0.002**.