\( X \) is a normally distributed random variable with mean 7 and standard deviation 10 . What is the probability that \( X \) is less than 27 ? 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 less than 27, we can use the properties of the normal distribution along with the \( 0.68-0.95-0.997 \) rule. 1. **Identify the mean and standard deviation:** - Mean (\( \mu \)) = 7 - Standard deviation (\( \sigma \)) = 10 2. **Calculate the z-score for \( X = 27 \):** The z-score is calculated using the formula: \[ z = \frac{X - \mu}{\sigma} \] Substituting the values: \[ z = \frac{27 - 7}{10} = \frac{20}{10} = 2 \] 3. **Use the z-score to find the probability:** A z-score of 2 corresponds to the area under the normal curve to the left of this z-score. According to the \( 0.68-0.95-0.997 \) rule: - Approximately 68% of the data falls within 1 standard deviation of the mean (between \( \mu - \sigma \) and \( \mu + \sigma \)). - Approximately 95% falls within 2 standard deviations (between \( \mu - 2\sigma \) and \( \mu + 2\sigma \)). - Approximately 99.7% falls within 3 standard deviations. Since \( z = 2 \) is within 2 standard deviations, we can conclude that approximately 95% of the data falls within this range. Therefore, the area to the left of \( z = 2 \) is: \[ P(X < 27) \approx 0.95 \] 4. **Final answer:** The probability that \( X \) is less than 27 is approximately \( 0.95 \). Thus, the final answer is: \[ \boxed{0.950} \]