A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poissondistribution to find the probability that during a randomly selected hour, the organization will receive fewerthan two calls.
A) 0.894
B) 0.106
C) 0.317
D) 0.087

Question

Quizplus · Accepted Answer

The Poisson distribution formula for the probability of observing k events in an interval is given by $P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average number of events per interval, and $k$ is the number of occurrences. For fewer than two calls, we calculate the probability of receiving 0 or 1 calls. With $\lambda = 0.55$, $P(0) = \frac{e^{-0.55} \cdot 0.55^0}{0!} = e^{-0.55}$, and $P(1) = \frac{e^{-0.55} \cdot 0.55^1}{1!} = 0.55e^{-0.55}$. Adding these probabilities gives the total probability of receiving fewer than two calls, which is approximately 0.894.

A Local Animal Rescue Organization Receives an Average of 0