\( X \) is a normally distributed random variable with mean 50 and standard deviation 21 . What is the probability that \( X \) is between 29 and 92 ? 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 between 29 and 92 using the \( 0.68-0.95-0.997 \) (68-95-99.7) rule, follow these steps: 1. **Identify the Mean and Standard Deviation:** - Mean (\( \mu \)) = 50 - Standard Deviation (\( \sigma \)) = 21 2. **Calculate the Z-scores:** - For \( X = 29 \): \[ Z_1 = \frac{29 - 50}{21} = -1 \] - For \( X = 92 \): \[ Z_2 = \frac{92 - 50}{21} = 2 \] 3. **Apply the Empirical Rule:** - The 68-95-99.7 rule states: - Approximately **68%** of the data falls within \( \pm 1\sigma \) of the mean. - Approximately **95%** falls within \( \pm 2\sigma \). - Since we're looking for the probability between \( -1\sigma \) and \( +2\sigma \): - **From \( -1\sigma \) to \( +1\sigma \)**: 68% - **From \( +1\sigma \) to \( +2\sigma \)**: Half of the remaining 27% (since 95% within \( \pm 2\sigma \)), which is 13.5% - **Total Probability:** \[ 68\% + 13.5\% = 81.5\% \] 4. **Express as a Decimal:** \[ 81.5\% = 0.815 \] Rounding to the nearest thousandth gives **0.816**. **Final Answer:** The probability is approximately **0.816**.