\( X \) is a normally distributed random variable with mean 19 and standard deviation 5 . What is the probability that \( X \) is between 4 and 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 the normally distributed random variable \( X \) with mean \( \mu = 19 \) and standard deviation \( \sigma = 5 \) falls between 4 and 29, we'll use the **68-95-99.7 rule** (also known as the empirical rule). ### Step-by-Step Calculation: 1. **Identify the Z-scores:** - For \( X = 4 \): \[ Z = \frac{4 - 19}{5} = \frac{-15}{5} = -3 \] - For \( X = 29 \): \[ Z = \frac{29 - 19}{5} = \frac{10}{5} = 2 \] 2. **Interpret the Z-scores using the empirical rule:** - **Within 1 standard deviation (\( \sigma \)):** \( \mu \pm 1\sigma \) covers **68%** of the data. - **Within 2 standard deviations (\( 2\sigma \)):** \( \mu \pm 2\sigma \) covers **95%** of the data. - **Within 3 standard deviations (\( 3\sigma \)):** \( \mu \pm 3\sigma \) covers **99.7%** of the data. 3. **Calculate the Probability between \( Z = -3 \) and \( Z = 2 \):** - **From \( Z = -3 \) to \( Z = 3 \):** Covers **99.7%**. - **From \( Z = -3 \) to \( Z = 2 \):** This interval is slightly less than \( Z = -3 \) to \( Z = 3 \) because it stops at \( Z = 2 \). - The area from \( Z = 2 \) to \( Z = 3 \) is approximately **2.35%** (half of the remaining 0.3% not covered within \( \pm 2\sigma \)). - Therefore, the probability between \( Z = -3 \) and \( Z = 2 \) is: \[ 99.7\% - 2.35\% \approx 97.35\% \] 4. **Express as a Decimal:** \[ 97.35\% \approx 0.9735 \] Rounding to the nearest thousandth gives **0.974**. ### Final Answer: The probability that \( X \) is between 4 and 29 is **0.974**.