1. **Problem statement:** Complaints about an Internet brokerage firm occur at a rate of 1.1 per day, and the number of complaints follows a Poisson distribution. We want to find the probability that the firm receives more than 4 complaints in a 3-day period. 2. **Formula and explanation:** For a Poisson distribution, the probability of observing $k$ events in a time period is given by: $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda$ is the expected number of events in that period. 3. **Calculate $\lambda$ for 3 days:** $$\lambda = 1.1 \times 3 = 3.3$$ 4. **Calculate probabilities for $k=0,1,2,3,4$:** - $P(X=0) = \frac{3.3^0 e^{-3.3}}{0!} = e^{-3.3} \approx 0.0367$ (3.7%) - $P(X=1) = \frac{3.3^1 e^{-3.3}}{1!} = 3.3 e^{-3.3} \approx 0.1211$ (12.1%) - $P(X=2) = \frac{3.3^2 e^{-3.3}}{2!} = \frac{10.89 e^{-3.3}}{2} \approx 0.1998$ (20.0%) - $P(X=3) = \frac{3.3^3 e^{-3.3}}{3!} = \frac{35.937 e^{-3.3}}{6} \approx 0.2197$ (22.0%) - $P(X=4) = \frac{3.3^4 e^{-3.3}}{4!} = \frac{118.5921 e^{-3.3}}{24} \approx 0.1813$ (18.1%) 5. **Calculate probability of receiving less than or equal to 4 complaints:** $$P(X \leq 4) = \sum_{k=0}^4 P(X=k) = 0.0367 + 0.1211 + 0.1998 + 0.2197 + 0.1813 = 0.7586$$ Rounded: 75.9% 6. **Calculate probability of receiving more than 4 complaints:** $$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.7586 = 0.2414$$ Rounded: 24.1% **Final answers:** (a) 3.7% (b) 12.1% (c) 20.0% (d) 22.0% (e) 18.1% (f) 75.9% (g) 24.1%