\( X \) is a normally distributed random variable with mean 58 and standard deviation 2 . What is the probability that \( X \) is greater than 60 ? 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 60, where \( X \) is normally distributed with a mean (\( \mu \)) of 58 and a standard deviation (\( \sigma \)) of 2, we can use the **68-95-99.7 rule** (also known as the empirical rule). 1. **Identify the Z-score:** \[ Z = \frac{X - \mu}{\sigma} = \frac{60 - 58}{2} = 1 \] So, 60 is 1 standard deviation above the mean. 2. **Apply the 68-95-99.7 Rule:** - **68%** of the data falls within **±1** standard deviation from the mean. - Therefore, the probability that \( X \) is **less than or equal to 60** is roughly **68% / 2 = 34%** above the mean. - The total area under the curve is 100%, so the probability that \( X \) is **greater than 60** is: \[ 50\% - 34\% = 16\% \] 3. **Convert to Decimal:** \[ 16\% = 0.16 \] **Final Answer:** The probability that \( X \) is greater than 60 is **0.160**.