Question

Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester. What is the probability that more than 3 students will have their automobiles stolen during the current semeste

Answer 1

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.

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:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` 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
`\mu = 7`

What is the probability that more than 3 students will have their automobiles stolen during the current semester?

This is:

`P(X > 3) = 1 - P(X \leq 3)`

In which

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)`

So

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-7)*7^(0))/((0)!) = 0.00091`

`P(X = 1) = (e^(-7)*7^(1))/((1)!) = 0.00638`

`P(X = 2) = (e^(-7)*7^(2))/((2)!) = 0.02234`

`P(X = 3) = (e^(-7)*7^(3))/((3)!) = 0.05213`

Then

`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`

`P(X > 3) = 1 - P(X \leq 3) = 1 - 0.08176 = 0.91824`

0.91824 = 91.824% probability that more than 3 students will have their automobiles stolen during the current semester.