A local fire station receives an average of 0.55 rescue calls per day. Use the Poisson distribution to find the probability that on a randomly selected day, the fire station will receive fewer than 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 a fixed interval is P(k; λ) = (e^(-λ) * λ^k) / k!, where λ is the average number of events (0.55 in this case) and k is the number of occurrences (here, we're interested in fewer than 2 calls, so we calculate for k=0 and k=1). For k=0: P(0; 0.55) = (e^(-0.55) * 0.55^0) / 0! = e^(-0.55). For k=1: P(1; 0.55) = (e^(-0.55) * 0.55^1) / 1! = 0.55 * e^(-0.55). Adding these probabilities gives the total probability of receiving fewer than two calls. The calculation matches closest to option A, indicating a high probability (0.894) of receiving fewer than two calls.

A Local Fire Station Receives an Average of 0