\( X \) is a normally distributed random variable with mean 37 and standard deviation 4 . What is the probability that \( X \) is less than 29 ? 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 29, where \( X \) is normally distributed with a mean (\( \mu \)) of 37 and a standard deviation (\( \sigma \)) of 4, we can use the \( 68-95-99.7 \) (empirical) rule. ### Step-by-Step Calculation: 1. **Find the Z-score:** \[ Z = \frac{X - \mu}{\sigma} = \frac{29 - 37}{4} = \frac{-8}{4} = -2 \] 2. **Interpret the Z-score using the \( 68-95-99.7 \) rule:** - The rule states: - Approximately **68%** of data falls within **±1** standard deviation. - Approximately **95%** within **±2** standard deviations. - Approximately **99.7%** within **±3** standard deviations. - A Z-score of **-2** corresponds to the lower 2 standard deviations from the mean. 3. **Determine the probability:** - Since \( Z = -2 \) is two standard deviations below the mean, the probability of \( X \) being less than 29 is roughly **2.5%**. 4. **Expressing the probability as a decimal:** \[ 0.025 \] ### Final Answer: The probability that \( X \) is less than 29 is **0.025**.