Question

A statistics professor finds that when he 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 6.

Answer 1

0.0662

Explanation:

The given problem indicates the Poisson experiment.

The pdf of Poisson experiment is as follow

P(X=x)=μ^x(e^-μ)/x!

Here, the average student visits in office hour=3.3=μ.

We have to find the probability that the number of student arrivals is 6 in a randomly selected office hour.

So, x=6 and μ=3.3

The probability that the number of student arrivals is 6 in a randomly selected office hour

P(X=6)=3.3^6(e^-3.3)/6!

P(X=6)=1291.468(.0369)/720

P(X=6)=47.6552/720

P(X=6)=0.0662.

Thus, the probability that the number of student arrivals is 6 in a randomly selected office hour is 6.62 or 0.0662.

Answer 2

`P(X=6)`

If we use the probability mass function we got:

`P(X=6) = (e^(-3.3) 3.3^6)/(6!)= 0.0662`

Explanation:

Previous concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

Solution to the problem

Let X the random variable that represent the number of students arrive at the office hour. We know that
`X \sim Poisson(\lambda=3.3)`

The probability mass function for the random variable is given by:

`f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...`

And f(x)=0 for other case.

For this distribution the expected value is the same parameter
`\lambda`

`E(X)=\mu =\lambda=3.3`

And we want this probability:

`P(X=6)`

If we use the probability mass function we got:

`P(X=6) = (e^(-3.3) 3.3^6)/(6!)= 0.0662`