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

0.0137

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)=(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 ((e^(-5)5^0)/(0!)+(e^(-5)5^1)/(1!)+...+(e^(-5)5^(10))/(10!)\right)=\\=1-e^(-5)\left((5^0)/(0!)+(5^1)/(1!)+...+(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