Question

A new phone answering system for a company is capable of handling sevenseven 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 fourfour every 10 minutes. If this incoming call distribution is what the analysts think it​ is, what is the probability that in a​ 10-minute period more calls will arrive than the system can​ handle?

Answer 1

Answer:

0.0166

Step-by-step explanation:

P = probability of more calls than system can handle = 1 – (probability of 0 calls + probability of 1 call + probability of 2 calls + probability of 3 calls + probability of 4 calls + probability of 5 calls)

Thus,

$P = 1 - (\frac{(2^0)\times e^{-2}}{0!} + \frac{(2^1)\times e^{-2}}{1!} + \frac{(2^2)\times(e^{-2})}{2!} + \frac{((2^3)\times(e^{-2}}{3!} + \frac{2^4\times(e^(-2))}{4!} + \frac{2^5\times e^{-2}}{{5!}}) = 0.0166$