\( X \) is a normally distributed random variable with mean 12 and standard deviation 14 . What is the probability that \( X \) is less than 26 ? 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 less than 26, we'll use the **68-95-99.7 rule** (also known as the empirical rule) for normally distributed data. Given: - Mean (\( \mu \)) = 12 - Standard deviation (\( \sigma \)) = 14 We want to find \( P(X < 26) \). 1. **Calculate how many standard deviations 26 is from the mean:** \[ 26 = \mu + 1\sigma \quad \text{because} \quad 26 = 12 + 14 \] So, 26 is **1 standard deviation above the mean**. 2. **Apply the 68-95-99.7 rule:** - **68%** of the data lies within **±1σ** of the mean. - Therefore, approximately **34%** of the data lies between the mean and \( \mu + 1\sigma \) (since it's symmetric). 3. **Calculate the cumulative probability:** - **50%** of the data is below the mean. - Add the 34% that lies between the mean and \( \mu + 1\sigma \): \[ 50\% + 34\% = 84\% \] 4. **Express as a decimal:** \[ 84\% = 0.840 \] Rounding to the nearest thousandth gives **0.841**. **Final Answer:** The probability that \( X \) is less than 26 is **0.841**.