Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A 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 visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
