Respuesta :
Answer:
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
Step-by-step explanation:
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 time interval.
In this problem, we have that:
2.83 rescues every eight hours.
What is the probability that the squad will have at most 2 calls in an hour?
This is
[tex]P = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
We have 2.83 rescues every 8 hours. So for an hour, we have [tex]\mu = \frac{2.83}{8} = 0.354[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019[/tex]
[tex]P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485[/tex]
[tex]P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440[/tex]
So
[tex]P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944[/tex]
There is a 99.44% probability that the squad will have at most 2 calls in an hour.
