1. **Problem Statement:** A bus arrives following a normal distribution with mean 8:55 AM and standard deviation 5 minutes. We want to find the probability the bus arrives earlier than 8:50 AM. 2. **Formula and Explanation:** For a normal distribution $X \sim N(\mu, \sigma)$, the probability $P(X < x)$ is found by standardizing: $$Z = \frac{X - \mu}{\sigma}$$ where $Z \sim N(0,1)$. 3. **Calculate Z-score:** Convert 8:50 AM to minutes after 8:00 AM: 50 minutes. Mean bus arrival time: 8:55 AM = 55 minutes. Standard deviation: 5 minutes. $$Z = \frac{50 - 55}{5} = \frac{-5}{5} = -1$$ 4. **Find Probability:** Using standard normal tables or a calculator: $$P(Z < -1) = 0.1587$$ **Final answer:** The probability that the bus arrives earlier than 8:50 AM is approximately 0.1587.