1. **Problem Statement:** We have an intersection with an average of 150 cars passing through every hour. We want to analyze the number of cars passing through in a 1-minute period. 2. **Part (a): Probability Distribution** Since the cars pass independently and the average rate is constant, the number of cars passing in a fixed interval follows a Poisson distribution. The Poisson distribution formula is: $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda$ is the average number of events (cars) in the interval, and $k$ is the number of events. 3. **Part (b): Mean and Standard Deviation** The average rate is 150 cars per hour, so for 1 minute: $$\lambda = \frac{150}{60} = 2.5$$ For a Poisson distribution: - Mean $\mu = \lambda = 2.5$ - Standard deviation $\sigma = \sqrt{\lambda} = \sqrt{2.5} \approx 1.58$ 4. **Part (c): Probability of Exactly 3 Cars and Unusualness** Calculate $P(X=3)$: $$P(X=3) = \frac{2.5^3 e^{-2.5}}{3!} = \frac{15.625 \times e^{-2.5}}{6}$$ Calculate $e^{-2.5} \approx 0.0821$: $$P(X=3) \approx \frac{15.625 \times 0.0821}{6} = \frac{1.282}{6} \approx 0.2137$$ Number of standard deviations from mean: $$z = \frac{3 - 2.5}{1.58} \approx 0.32$$ Since $z$ is small and probability is relatively high, having exactly 3 cars is not unusual. 5. **Part (d): Probability of Between 5 and 10 Cars** Calculate $P(5 \leq X \leq 10) = \sum_{k=5}^{10} P(X=k)$. Using Poisson probabilities: - $P(X=5) = \frac{2.5^5 e^{-2.5}}{5!} \approx 0.0668$ - $P(X=6) \approx 0.0278$ - $P(X=7) \approx 0.0099$ - $P(X=8) \approx 0.0031$ - $P(X=9) \approx 0.0009$ - $P(X=10) \approx 0.0002$ Sum: $$0.0668 + 0.0278 + 0.0099 + 0.0031 + 0.0009 + 0.0002 = 0.1087$$ This probability is about 10.87%, which is low but not extremely rare, so it is somewhat unusual but not highly unlikely. **Final answers:** - (a) Poisson distribution with $\lambda=2.5$ cars per minute. - (b) Mean = 2.5, Standard deviation $\approx 1.58$. - (c) Probability of exactly 3 cars $\approx 0.214$, not unusual. - (d) Probability between 5 and 10 cars $\approx 0.109$, somewhat unusual.