Answer :
Answer:
0.713 = 71.3% probability that the county office will get more than 0 calls in a 15 minute period.
Step-by-step explanation:
We have the mean during a time-period, which means that the Poisson distribution is used to solve this question.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
A county office gets an average of 10 calls in a 2 hour time period.
10 calls each 120 minutes, which means that the mean for n minutes is:
[tex]\mu = \frac{10n}{120} = \frac{n}{12}[/tex]
15 minute period:
This means that [tex]n = 15, \mu = \frac{15}{12} = 1.25[/tex]
What is the probability that the county office will get more than 0 calls in a 15 minute period?
This is:
[tex]P(X > 0) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.25}*1.25^{0}}{(0)!} = 0.287[/tex]
So
[tex]P(X > 0) = 1 - P(X = 0) = 1 - 0.287 = 0.713[/tex]
0.713 = 71.3% probability that the county office will get more than 0 calls in a 15 minute period.