Question

The university police department must write, on average, five tickets per day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 7.5. Find the probability that fewer than three tickets are written on a randomly selected day.

Answer 1

0.0204 = 2.04% probability that fewer than three tickets are written on a randomly selected day.

Explanation:

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.

Poisson distribution with a mean of 7.5.

This means that
`\mu = 7.5`.

Find the probability that fewer than three tickets are written on a randomly selected day.

This is:

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

So

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

`P(X = 0) = (e^(-7.5)*(7.5)^(0))/((0)!) = 0.0006`

`P(X = 1) = (e^(-7.5)*(7.5)^(1))/((1)!) = 0.0042`

`P(X = 2) = (e^(-7.5)*(7.5)^(2))/((2)!) = 0.0156`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0006 + 0.0042 + 0.0156 = 0.0204`

0.0204 = 2.04% probability that fewer than three tickets are written on a randomly selected day.