Question

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.

Answer 1

Answer:

0.0137

Step-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

$\bf P(X=k)=\frac{e^{-5}5^k}{k!}$

If the phone-answering system is capable of handling ten calls every 10 minutes, we want to find

P(X>10), or the equivalent 1 - P(X≤ 10).  

But

1 - P(X≤ 10)= 1 -(P(X=0)+P(X=1)+...+P(X=10)) =

$\bf 1-\left (\frac{e^{-5}5^0}{0!}+\frac{e^{-5}5^1}{1!}+...+\frac{e^{-5}5^{10}}{10!}\right)=\\=1-e^{-5}\left(\frac{5^0}{0!}+\frac{5^1}{1!}+...+\frac{5^{10}}{10!}\right)=1-0.9863=0.0137$

So, the probability that in a 10-minute period more calls will arrive than the system can​ handle is 0.0137