Question

A statistics professor finds that when she schedules an office hour for student help, an average of 3.3 students arrive. Find the probability that in a randomly selected office hour, the number of student arrivals is 3.

Answer 1

0.2209 = 22.09% probability that in a randomly selected office hour, the number of student arrivals is 3.

Explanation:

We have the mean during an interval, so the Poisson distribution is used.

Poisson distribution:

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.

A statistics professor finds that when she schedules an office hour for student help, an average of 3.3 students arrive.

This means that
`\mu = 3.3`

Find the probability that in a randomly selected office hour, the number of student arrivals is 3.

This is P(X = 3). So

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

`P(X = 3) = (e^(-3.3)*3.3^(3))/((3)!) = 0.2209`

0.2209 = 22.09% probability that in a randomly selected office hour, the number of student arrivals is 3.