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

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.00091$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.00638$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.02234$

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.05213$

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.