\( X \) is a normally distributed random variable with mean 62 and standard deviation 10 . What is the probability that \( X \) is between 32 and 72 ? 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 32 and 72, we'll use the Empirical Rule (68-95-99.7 rule) for normally distributed variables. 1. **Mean and Standard Deviation:** - Mean (\( \mu \)) = 62 - Standard Deviation (\( \sigma \)) = 10 2. **Identify the Z-scores:** - 32 is \( 3\sigma \) below the mean: \( 62 - 3 \times 10 = 32 \) - 72 is \( 1\sigma \) above the mean: \( 62 + 1 \times 10 = 72 \) 3. **Apply the Empirical Rule:** - **Within \( 1\sigma \) (52 to 72):** Approximately 68% of the data falls here. - **Within \( 3\sigma \) (32 to 92):** Approximately 99.7% of the data falls here. 4. **Calculate the Probability:** - **Probability \( X \leq 72 \):** Approximately 84% (since 50% is below the mean, plus 34% for the first \( \sigma \)). - **Probability \( X < 32 \):** Approximately 0.15% (since it's 3 standard deviations below the mean). 5. **Final Probability:** \[ P(32 \leq X \leq 72) = P(X \leq 72) - P(X < 32) \approx 84\% - 0.15\% = 83.85\% \] Rounded to the nearest thousandth as a decimal: \[ 0.839 \] **Answer:** The probability is approximately 0.839.