Respuesta :
Answer:
The expected number of calls in one hour is 30.
The probability of three calls in five minutes is [tex]P(3) \approx0.2138[/tex]
The probability of no calls in a five-minutes is [tex]P(0) \approx 0.0821[/tex]
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:
[tex]\frac{1 \:call}{2 \:minutes} \cdot 60 \:minutes=30 \:calls[/tex]
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
[tex]P(x)=\frac{e^{-\mu}\mu^x}{x!}[/tex]
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
[tex]\mu=\frac{1 \:call}{2 \:minutes} \cdot 5 \:minutes=2.5 \:calls[/tex]
Evaluate the formula for k = 3 and μ = 2.5
[tex]P(3)=\frac{e^{-2.5}2.5^3}{3!}\\P(3) \approx0.2138[/tex]
- 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
[tex]P(0)=\frac{e^{-2.5}2.5^0}{0!}\\P(0) \approx 0.0821[/tex]
