Answer:
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)!)](https://img.qammunity.org/2022/formulas/mathematics/college/fc9bfg9bauetugxxr4o8egdqz83cs0jk74.png)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
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](https://img.qammunity.org/2022/formulas/mathematics/college/j872xjuat15oqa41e3k95lanld1s3jaoov.png)
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)!)](https://img.qammunity.org/2022/formulas/mathematics/college/fc9bfg9bauetugxxr4o8egdqz83cs0jk74.png)
![P(X = 3) = (e^(-3.3)*3.3^(3))/((3)!) = 0.2209](https://img.qammunity.org/2022/formulas/mathematics/college/r5cnzaxipuf4gk2p8igfhabhrysw415m0c.png)
0.2209 = 22.09% probability that in a randomly selected office hour, the number of student arrivals is 3.