Respuesta :
Answer:
16.81% probability that, for any day, the number of special orders sent out will be exactly 4
Step-by-step explanation:
The only information that we have is a mean during an interval. So we use the Poisson distribution solve this question.
We have that the probability of exactly x events is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}[/tex]
In which [tex]\mu[/tex] is the mean.
The auto parts department of an automotive dealership sends out a mean of 5.2 special orders daily.
This means that [tex]\mu = 5.2[/tex]
What is the probability that, for any day, the number of special orders sent out will be exactly 4?
This is P(X = 4).
[tex]P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}[/tex]
[tex]P(X = 4) = \frac{e^{-5.2}*(5.2)^{4}}{4!} = 0.1681[/tex]
16.81% probability that, for any day, the number of special orders sent out will be exactly 4
