Reply With QuoteYour answer seems logical... I get that t ≥ 41 sec...
Is that not right? What is the right answer then?
My first post! Anyway I just need a bit of help with a question that I've come across in a textbook which is as follows:
Cars arrive at a toll both according to a Poisson process with mean 80 cars per hour. How long can an attendent’s phone call last if the probability is at least .4 that no cars arrive?
Now I just thought this would simply be found by solving for
P(Y=0)≥0.4 with lambda = 4t/3 (working in minutes instead of hours)
So ((4/3)^0).e^-4t/3)/0! ≥0.4
But this is not the answer, and I do not know why it isn’t.
Thanks!
Your answer seems logical... I get that t ≥ 41 sec...
Is that not right? What is the right answer then?
Last edited by Riverdale27; 02-11-2010 at 07:58 AM.
Well apparently the answer is actually P(Y=0) = 1 - e^(-4t/3) ≥ 0.4
which is .383 or 23 seconds.
This makes no sense to me. Mistake in the book? It is singled out as a hard question, and if the answer were as straight forward as I think it is I doubt it would qualify for a hard question. So I guess there is some reason I cannot fathom why this isn't a straight forward poisson distribution question.
Thanks for the interest.
It has to do with the relationship between the Poisson and exponential distributions.
Most texts show this relationship, but basically it is that if arrivals follow a Poisson distribution (discrete), then the time between arrivals will follow an exponential distribution (continuous).
The 1 - e^(-4t/3) solution is actually the cumulative exponential distribution.
The text doesn't really talk about that, is there anywhere I could find out more about cumulative poisson distributions? Because I still don't follow as to why that is the answer.
The Poisson is a discrete probability distribution used to model arrivals (among other things). Therefore, it can give you a probability of Pr(X=x) number of arrivals over some time interval. The cumulative Poisson can give you the probability of Pr(X<=x) arrivals over some time period.
It is shown in most texts that if arrivals follow a Poisson distribution, then the time between arrivals will be exponentially distributed. The exponential distribution is a continuous distribution (not discrete).
Since the question was "how long" and not "how many", it is referring to the time between arrivals (exponentially distributed). The exponential distribution for Pr(0< x < t) = 1 - e^(-lambda(t)).
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