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

Answer:

$P(X=6)$

If we use the probability mass function we got:

$P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662$

Step-by-step 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:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

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:

$P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662$

Answer 2

Answer:

0.0662

Step-by-step 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.