Answer:
0.67% probability he will have to shut down after this month
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.
On average sells 8.9 machines per month.
So [tex]\mu = 8.9[/tex]
Using the Poisson distribution, what is the probability he will have to shut down after this month
If he sells less than 3 machines.
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-8.9}*8.9^{0}}{(0)!} = 0.0001[/tex]
[tex]P(X = 1) = \frac{e^{-8.9}*8.9^{1}}{(1)!} = 0.0012[/tex]
[tex]P(X = 2) = \frac{e^{-8.9}*8.9^{2}}{(2)!} = 0.0054[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0001 + 0.0012 + 0.0054 = 0.0067[/tex]
0.67% probability he will have to shut down after this month
