\( X \) is a normally distributed random variable with mean 84 and standard deviation 6 . What is the probability that \( X \) is less than 66 ? 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 66, given that \( X \) is normally distributed with a mean (\( \mu \)) of 84 and a standard deviation (\( \sigma \)) of 6, we'll use the \( 68-95-99.7 \) rule (also known as the Empirical Rule). ### Step-by-Step Calculation: 1. **Determine the Z-Score:** The Z-score measures how many standard deviations an element is from the mean. It's calculated as: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 66 \): \[ Z = \frac{66 - 84}{6} = \frac{-18}{6} = -3 \] 2. **Apply the 68-95-99.7 Rule:** - **68%** of the data lies within **±1** standard deviation from the mean. - **95%** within **±2** standard deviations. - **99.7%** within **±3** standard deviations. Since \( Z = -3 \) is exactly 3 standard deviations below the mean, the probability that \( X \) is less than 66 corresponds to the extreme end of the distribution. The rule tells us that: \[ P(\mu - 3\sigma < X < \mu + 3\sigma) = 0.997 \] Therefore, the probability that \( X \) is **less than** \( \mu - 3\sigma \) (which is 66) is: \[ P(X < 66) = \frac{1 - 0.997}{2} = \frac{0.003}{2} = 0.0015 \] Rounding to the nearest thousandth: \[ P(X < 66) \approx 0.001 \] ### Final Answer: \[ \boxed{0.001} \]