Question

During the period of time that a local university takes phone-in registrations, calls come in

Answer 1

Answer:

a) The expected number of calls in one hour is 30.

b) There is a 21.38% probability of three calls in five minutes.

c) There is an 8.2% probability of no calls in a five minute period.

Step-by-step explanation:

In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.

Poisson probability distribution

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

a. What is the expected number of calls in one hour?

Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.

b. What is the probability of three calls in five minutes?

Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that $\mu = 2.5$. We want to find P(X = 3).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 3) = \frac{e^{-2.5}*(2.5)^{3}}{(3)!} = 0.2138$

There is a 21.38% probability of three calls in five minutes.

c. What is the probability of no calls in a five-minute period?

This is P(X = 0) with $\mu = 2.5$.

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2.5}*(2.5)^{0}}{(0)!} = 0.0820$

There is an 8.2% probability of no calls in a five minute period.