Respuesta :
Answer:
0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.
Step-by-step explanation:
We have only 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.
Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester.
This means that [tex]\mu = 7[/tex]
What is the probability that more than 3 students will have their automobiles stolen during the current semester?
This is:
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.00091[/tex]
[tex]P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.00638[/tex]
[tex]P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.02234[/tex]
[tex]P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.05213[/tex]
Then
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.00091 + 0.00638 + 0.02234 + 0.05213 = 0.08176 [/tex]
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.08176 = 0.91824[/tex]
0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.
