Respuesta :
Answer:
0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.
Step-by-step explanation:
We have the mean, 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 mean of 3.5 customers arrive hourly at the drive-through window.
This means that [tex]\mu = 3.5[/tex]
What is the probability that, in any hour, more than 5 customers will arrive?
This is:
[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]
In which
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
Then
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.5}*3.5^{0}}{(0)!} = 0.0302[/tex]
[tex]P(X = 1) = \frac{e^{-3.5}*3.5^{1}}{(1)!} = 0.1057[/tex]
[tex]P(X = 2) = \frac{e^{-3.5}*3.5^{2}}{(2)!} = 0.1850[/tex]
[tex]P(X = 3) = \frac{e^{-3.5}*3.5^{3}}{(3)!} = 0.2158[/tex]
[tex]P(X = 4) = \frac{e^{-3.5}*3.5^{4}}{(4)!} = 0.1888[/tex]
[tex]P(X = 5) = \frac{e^{-3.5}*3.5^{5}}{(5)!} = 0.1322[/tex]
Finally
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577[/tex]
[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423[/tex]
0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.
