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Poisson Distribution

Hey! So I'm pretty sure this question is asking me to use poisson to solve it, but it doesn't explicitly say that, so if I'm wrong let me know!

Trains arrive at a station at the rate of 25 per hour.
a) What is the probability that the time between trains arriving is 6 minutes orless?b) The train company wants to state that waiting times (in minutes) betweentrains have been targeted for improvement. It fails to meet its target 40% of thetime. What target has it set itself?
c) A train enthusiast notes that a goods train arrives, on average, once an hour.

So like I said, I think its poisson, so would the mean be 25 divided by 60? Using that I got 0.999 for part a, because its asking you you to find P(X<=6) right? To get it I did P(0) + P(1) + P(2) + P(3) etc up to P(6). Sound right?

I'm kinda stuck b and c though...

Any help would be much appreciated! I just need to know the methods :biggrin:
Reply 1
Original post by JaketheSnake72
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It is Poisson, yes. So is 25 is in one hour, then it's 2.5 in 6 minutes. So you define Y ~ Po(2.5) and then use that for part (a).
Original post by Zacken
It is Poisson, yes. So is 25 is in one hour, then it's 2.5 in 6 minutes. So you define Y ~ Po(2.5) and then use that for part (a).


Thank! So a is 0.985? Then what's the method for b? I'm not even sure where to start...
Reply 3
Original post by JaketheSnake72
Thank! So a is 0.985? Then what's the method for b? I'm not even sure where to start...


I don't think A is correct. "Trains arriving less than 6 minutes together" means that in a time period of 6 minutes you want there to be at least 1 train arriving. So Y ~ Po(2.5) then P(Y >= 1).

For (b). If you want there to be train arriving in less than k minutes together means that in a time period of k minutes (so mean is now what? If it's 25 in 60 minutes, how much is it in k minutes?) Then you need P(X < 1) = 0.4. So What does k have to be?
Original post by JaketheSnake72
Hey! So I'm pretty sure this question is asking me to use poisson to solve it, but it doesn't explicitly say that, so if I'm wrong let me know!

Trains arrive at a station at the rate of 25 per hour.
a) What is the probability that the time between trains arriving is 6 minutes orless?b) The train company wants to state that waiting times (in minutes) betweentrains have been targeted for improvement. It fails to meet its target 40% of thetime. What target has it set itself?
c) A train enthusiast notes that a goods train arrives, on average, once an hour.

So like I said, I think its poisson, so would the mean be 25 divided by 60? Using that I got 0.999 for part a, because its asking you you to find P(X<=6) right? To get it I did P(0) + P(1) + P(2) + P(3) etc up to P(6). Sound right?

I'm kinda stuck b and c though...

Any help would be much appreciated! I just need to know the methods :biggrin:


If arrivals are governed by a Poisson process with rate λ\lambda then the inter-arrival times are distributed as an exponential distribution with mean 1λ\frac{1}{\lambda}.

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