A new phone-answering system installed by a certain utility company is capable of handling ten calls every 10 minutes. Prior to installing the new system, company analysts determined that the incoming calls to the system are Poisson distributed with a mean equal to five every 10 minutes. If the analysts are correct about this incoming call distribution, what is the probability that in a 10-minute period more calls will arrive than the system can handle? Based on this probability, comment on the adequacy of the new answering system. The probability that more calls will arrive than the system can handle is nothing.
Accepted Solution
A:
Answer:0.0137Step-by-step explanation:Let X be the random variable that measures the number of incoming calls every ten minutes.If the incoming calls to the system are Poisson distributed with a mean equal to 5 every 10 minutes, then the probability that there are k incoming calls in 10 minutes is[tex] \bf P(X=k)=\frac{e^{-5}5^k}{k!}[/tex]If the phone-answering system is capable of handling ten calls every 10 minutes, we want to findP(X>10), or the equivalent 1 - P(X≤ 10). But1 - P(X≤ 10)= 1 -(P(X=0)+P(X=1)+...+P(X=10)) =[tex] \bf 1-\left (\frac{e^{-5}5^0}{0!}+\frac{e^{-5}5^1}{1!}+...+\frac{e^{-5}5^{10}}{10!}\right)=\\=1-e^{-5}\left(\frac{5^0}{0!}+\frac{5^1}{1!}+...+\frac{5^{10}}{10!}\right)=1-0.9863=0.0137[/tex]So, the probability that in a 10-minute period more calls will arrive than the system can handle is 0.0137