Poisson Distribution - number of vehicles at a toll bridge

Hi,

Question:

The number of vehicles arriving at a toll bridge during a 5-minute period can be modelled by a Poisson distribution with mean 3.6.

a) State the value for the standard deviation of the number of vehicles arriving at the toll during a 5-minute period.

S.D. = sqrt. (3.6)
S.D. = 1.90 (to 3 s.f.)

b) Find:
i) the probability that at least three vehicles arrive in a 5-minute period.

Using a cumulative Poisson distribution table:

P(X is greater than or equal to 3) = 1 - P(X is less than or equal to 2)
P(X is greater than or equal to 3) = 1 - 0.3027
P(X is greater than or equal to 3) = 0.6973

ii) the probability that at least three vehicles arrive in each of three successive 5-minute periods.

I'm stuck on part (ii). I think this is binomial but I'm not sure how to set it up. I did X~B(15, 0.6973) but it didn't work out correctly when I did 1 - (P(X=0) + P(X=1) + P(X=2)).

I know the answer is 0.339. I managed to get this by doing 0.6973^3 but I don't understand why this gives the answer. Would anyone be able to explain?

Thank you very much.

The Poisson distribution has a property called independent increments. That is, events in disjoint intervals are statistically independent. Put another way, the number of events that occur in one interval does not depend on the number in a previous interval or a future interval.

Thanks for the reply. I understand what you are saying but I still don't have a clue as to how to start part (ii). Any tips? :?

So is it 0.6973^3? I'm not sure how what you said fits in with that. If the events are "independent" then you wouldn't multiply 0.6973 x 0.6973 x 0.6973. Is that right?

I'm totally stumped about part (ii). Please, any more help is greatly appreciated.

The probability that at least 3 vehicles arrive in an arbitrary 5 minute period is .6973, as you noted. So in 3 independent or successive periods, the probability is indeed .6973^3.

Thanks royhaas, I get it now. Thanks. So because the events are independent, you can just do 0.6973^3.

Just 1 last thing I'm wondering about. In this similar question you answered for me:

viewtopic.php?f=12&t=30783

b) What is the probability that 10 consecutive days will include two or more on which the demand is zero?

Click to expand...

Why did we have to use binomial for that question about car hire? Whereas for this one about the toll bridge we didn't?

Is it because for the question about car hire, 10 days represents 1 whole period, but in the question about the toll bridge, we have 3 separate periods, each 5-minutes?

Thanks, I just want to make sure I understand.

It's the "2 or more" out of 10 days that requires the use of the binomial. Otherwise, which two days see a zero demand? In the current Poisson problem, you are using a binomial when you cube the probability. It's the probability of "three successes and zero failures" for a binomial with n=3.