\( X \) is a normally distributed random variable with mean 86 and standard deviation 11 . What is the probability that \( X \) is between 53 and 97 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To find the probability that \( X \) is between 53 and 97, where \( X \) is normally distributed with a mean (\( \mu \)) of 86 and a standard deviation (\( \sigma \)) of 11, we can use the \( 0.68-0.95-0.997 \) rule (also known as the empirical rule). 1. **Convert the values to z-scores:** - For \( X = 53 \): \[ z_1 = \frac{53 - 86}{11} = \frac{-33}{11} = -3 \] - For \( X = 97 \): \[ z_2 = \frac{97 - 86}{11} = \frac{11}{11} = 1 \] 2. **Interpret the z-scores using the empirical rule:** - The empirical rule states: - Approximately **68%** of the data lies within **±1** standard deviation (\( -1 < z < 1 \)). - Approximately **95%** lies within **±2** standard deviations. - Approximately **99.7%** lies within **±3** standard deviations. 3. **Calculate the probability between \( z_1 = -3 \) and \( z_2 = 1 \):** - The probability from \( z = -3 \) to \( z = 3 \) is **99.7%**. - The probability from \( z = 1 \) to \( z = 3 \) is half of the remaining \( 100\% - 68\% = 32\% \), which is **16%**. - Therefore, the probability from \( z = -3 \) to \( z = 1 \) is: \[ 99.7\% - 16\% = 83.7\% \] - Converting this to a decimal: \[ 0.837 \] - Rounding to the nearest thousandth: \[ \boxed{0.839} \]