\( X \) is a normally distributed random variable with mean 7 and standard deviation 22 . What is the probability that \( X \) is between 51 and 73 ? 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 51 and 73 using the \( 68-95-99.7 \) rule, follow these steps: 1. **Identify the Mean (\( \mu \)) and Standard Deviation (\( \sigma \)):** - Given \( \mu = 7 \) and \( \sigma = 22 \). 2. **Calculate the Z-scores for 51 and 73:** - For \( X = 51 \): \[ z_1 = \frac{51 - 7}{22} = \frac{44}{22} = 2 \] - For \( X = 73 \): \[ z_2 = \frac{73 - 7}{22} = \frac{66}{22} = 3 \] 3. **Apply the \( 68-95-99.7 \) Rule:** - The rule states: - **68%** of data lies within \( \mu \pm 1\sigma \). - **95%** within \( \mu \pm 2\sigma \). - **99.7%** within \( \mu \pm 3\sigma \). - To find \( P(2 \leq Z \leq 3) \): - The probability within \( 2\sigma \) is **95%**. - The probability within \( 3\sigma \) is **99.7%**. - Therefore, the probability between \( 2\sigma \) and \( 3\sigma \) is: \[ 99.7\% - 95\% = 4.7\% \] 4. **Express the Probability as a Decimal:** \[ 4.7\% = 0.047 \] **Final Answer:** The probability is 0.047