SOLUTION: This is a Poisson probability distribution problem. I thought it was easy but I can't get the answer right. I hope you'll find time to help me. Thanks. A Statistics Professor fi

Question 55312: This is a Poisson probability distribution problem. I thought it was easy but I can't get the answer right. I hope you'll find time to help me. Thanks.
A Statistics Professor finds that when she schedules an office hour for student help, an average of 2 students arrive. Find the the probability that in a randomly selected hour, the number of student arrival is 5. Give the answer as a decimal value with 4 places of precision and no leading 0.
P(x) = 2/8 = 0.25 ( I used 8 as I thought it meant 8 office hours)
P(x) = (0.25)^2 * 2.71828^-0.25/5!
= 0.0625 * .779/120
= .0004057 (or should it be .4057)
Again, thanks
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I'm using the following formula:
P(n,T) =(mu*T)^n][e^(-muT)]/n!
Your mu is 2; your T = 1 hr. ; your n=5

Actual arrivals in one hour were 2.
P(5 arrivals in one hour) = (2(1))^5 e^(-2(1))/5!
=[32e^-2]/120
=[32]/[120e^2]
=0.0361
Cheers,
Stan H.

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