EE. 3 Find probabilities using the normal distribution I SSW \( X \) is a normally distributed random variable with mean 25 and standard deviation 2 . What is the probability that \( X \) is greater than 23 ? 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 23 using the **68-95-99.7 rule** (also known as the empirical rule), follow these steps: ### Given: - Mean (\( \mu \)) = 25 - Standard Deviation (\( \sigma \)) = 2 - We need to find \( P(X > 23) \) ### Steps: 1. **Identify the Z-Score:** - \( 23 \) is \( 1 \sigma \) below the mean. - \( 23 = \mu - 1\sigma = 25 - 2 = 23 \) 2. **Apply the 68-95-99.7 Rule:** - **68%** of the data lies within \( \mu \pm 1\sigma \), which is between 23 and 27. - This means that **16%** of the data lies below 23 (since \( 100\% - 68\% = 32\% \) and half of that is below 23). - Therefore, the probability that \( X \) is greater than 23 is: \[ P(X > 23) = 1 - P(X < 23) = 1 - 0.16 = 0.84 \] ### Final Answer: The probability that \( X \) is greater than 23 is **0.840** (rounded to the nearest thousandth). **Answer:** The probability that \( X \) is greater than 23 is **0.840**.