Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Ask your questions and receive precise answers from experienced professionals across different disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
0.2209 = 22.09% probability that in a randomly selected office hour, the number of student arrivals is 3.
Step-by-step 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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] 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 [tex]\mu = 3.3[/tex]
Find the probability that in a randomly selected office hour, the number of student arrivals is 3.
This is P(X = 3). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-3.3}*3.3^{3}}{(3)!} = 0.2209[/tex]
0.2209 = 22.09% probability that in a randomly selected office hour, the number of student arrivals is 3.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
