\( X \) is a normally distributed random variable with mean 77 and standard deviation 11 . What is the probability that \( X \) is between 55 and 99 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

Given that \( X \) is normally distributed with a mean (\( \mu \)) of 77 and a standard deviation (\( \sigma \)) of 11, we want to find the probability that \( X \) lies between 55 and 99. First, determine how many standard deviations away from the mean the values 55 and 99 are: \[ \frac{55 - 77}{11} = -2 \quad \text{and} \quad \frac{99 - 77}{11} = 2 \] This means 55 is 2 standard deviations below the mean and 99 is 2 standard deviations above the mean. According to the \( 0.68-0.95-0.997 \) rule (also known as the empirical rule): - Approximately **68%** of the data lies within **1** standard deviation of the mean. - Approximately **95%** of the data lies within **2** standard deviations of the mean. - Approximately **99.7%** of the data lies within **3** standard deviations of the mean. Since both 55 and 99 are exactly 2 standard deviations away from the mean, the probability that \( X \) falls between 55 and 99 is approximately **95%**. Expressed as a decimal and rounded to the nearest thousandth: \[ \boxed{0.950} \]