Question

During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes. What is the expected number of calls in one hour? What is the probability of three calls in five minutes (to 4 decimals)? What is the probability of no calls in a five-minute period (to 4 decimals)?

Answer 1

Answer:

The expected number of calls in one hour is 30.

The probability of three calls in five minutes is $P(3) \approx0.2138$

The probability of no calls in a five-minutes is $P(0) \approx 0.0821$

Step-by-step explanation:

We know that the calls come in at the rate of one every two minutes. To find the expected number of calls in one hour you need to:

$\frac{1 \:call}{2 \:minutes} \cdot 60 \:minutes=30 \:calls$

To find the probability of these events:

• three calls in five minutes
• no calls in a five-minutes

We are going to use the Poisson distribution because is used when it is desirable to determine the probability of obtaining x occurrences over an interval of time like in this situation.

The Poisson probability function is defined by

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

where

x = 0, 1, 2, 3 ...

e = 2.71828

μ = mean number of successes in the given time interval or region of space

• To find the probability of three calls in five minutes you need to:

Find μ

$\mu=\frac{1 \:call}{2 \:minutes} \cdot 5 \:minutes=2.5 \:calls$

Evaluate the formula for k = 3 and μ = 2.5

$P(3)=\frac{e^{-2.5}2.5^3}{3!}\\P(3) \approx0.2138$

• To find the probability of no calls in a five-minute period you need to:

Evaluate the formula for  k = 0 and μ = 2.5

$P(0)=\frac{e^{-2.5}2.5^0}{0!}\\P(0) \approx 0.0821$