Emergency 911 calls to a small municipality in Idaho come in at the rate of one every two minutes. Assume that the number of 911 calls is a random variable that can be described by the Poisson distribution. (a) What is the expected number of 911 calls in one hour? per hour (b) What is the probability of three 911 calls in five minutes? If required, round your answer to four decimal places. (c) What is the probability of no 911 calls during a five-minute period? If required, round your answer to four decimal places.

Answer: a) 30, b) 0.2138, c) 0.0820Step-by-step explanation:Since we have given that X be the number of calls in 1 hour = 60 minutesRate of one call every two minutes.[tex]\lambda=\dfrac{1}{2}[/tex](a) What is the expected number of 911 calls in one hour? [tex]\lambda=\dfrac{1}{2}\times 60=30[/tex](b) What is the probability of three 911 calls in five minutes? If required, round your answer to four decimal places. [tex]\lambda=\dfrac{1}{2}\times 5=2.5[/tex]Number of calls = 3So, P(X=3) is given by[tex]P(X=3)=\dfrac{e^{-\lambda}\lambda^x}{x!}\\\\P(X=3)=\dfrac{e^{-2.5}\times (2.5)^3}{3!}\\\\P(X=3)=0.2138[/tex](c) What is the probability of no 911 calls during a five-minute period? If required, round your answer to four decimal places.[tex]P(X=0)=\dfrac{e^{-2.5}(2.5)^0}{0!}=0.0820[/tex]Hence, a) 30, b) 0.2138, c) 0.0820