Respuesta :
Answer:
0.1944 = 19.44% probability that, for any day, the number of special orders sent out will be exactly 4
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 interval.
The auto parts department of an automotive dealership sends out a mean of 3.8 special orders daily.
This means that [tex]\mu = 3.8[/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^{-3.8}*(3.8)^{4}}{(4)!} = 0.1944[/tex]
0.1944 = 19.44% probability that, for any day, the number of special orders sent out will be exactly 4
