Respuesta :
Answer:
0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3
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.
The auto parts department of an automotive dealership sends out a mean of 6.3 special orders daily.
This means that [tex]\mu = 6.3[/tex]
What is the probability that, for any day, the number of special orders sent out will be exactly 3?
This is P(X = 3). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-6.3}*6.3^{3}}{(3)!} = 0.0765[/tex]
0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3
