\( X \) is a normally distributed random variable with mean 73 and standard deviation 18 . What is the probability that \( X \) is less than 55 ? 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 less than 55, given that \( X \) is normally distributed with a mean (\( \mu \)) of 73 and a standard deviation (\( \sigma \)) of 18, we can use the **68-95-99.7 rule**. This rule provides a quick way to estimate probabilities for normally distributed data based on standard deviations from the mean. ### Step-by-Step Solution: 1. **Calculate the Z-score:** \[ Z = \frac{X - \mu}{\sigma} = \frac{55 - 73}{18} = \frac{-18}{18} = -1 \] A Z-score of -1 indicates that 55 is one standard deviation below the mean. 2. **Apply the 68-95-99.7 Rule:** - **68%** of the data falls within ±1 standard deviation from the mean. - Therefore, **32%** of the data falls outside this range (100% - 68%). - Since the normal distribution is symmetric, half of this 32% lies below \( \mu - 1\sigma \) and half above \( \mu + 1\sigma \). 3. **Calculate the Probability:** \[ P(X < 55) \approx \frac{32\%}{2} = 16\% \] Expressed as a decimal, this probability is **0.16**. ### Final Answer: The probability that \( X \) is less than 55 is **0.16**.