Question

A statistics professor finds that when he schedules an office hour for student help, an average of 1 students arrive. Find the probability that in a randomly selected office hour, the number of student arrivals is 3.

Answer 1

Answer:

6.13% probability that in a randomly selected office hour, the number of student arrivals is 3.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

A statistics professor finds that when he schedules an office hour for student help, an average of 1 students arrive.

This means that $\mu = 1$.

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

This is P(X = 3). So

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

$P(X = 3) = \frac{e^{-1}*(1)^{3}}{(3)!} = 0.0613$

6.13% probability that in a randomly selected office hour, the number of student arrivals is 3.