During rush hours at a Paris metro station, trains arrive according to a Poisson

Question

During rush hours at a Paris metro station, trains arrive according to a Poisson distribution with an expected number of ten trains per 60 minute period.

(a) What is the probabilty mass function of train arrivals in a period of length t minutes?

(b) What is the probability that two trains will arrive in a three minute period?

(c) What is the probabilty that no trains will arrive in a ten minute period?

(d) Find the probability that at least one train will arrive in a period of length t minutes and use this to compute how long is needed to be 95% sure that a train will arrive?

Explanation / Answer

= 10*t /60 = t/6

a) p(x,t) = x e- / x!

= (t/6)x e-(t/6) / x!

b) p(2 ,3) = (3/6)2 e-(3/6) / 2! = 0.075816

c) p(0,10) = (10/6)0 e-(10/6) / 0! = 0.18888

d) p(x>=1 ,t ) = 1 - p(0,t)

= 1 - (t/6)0 e-(t/6) / 0!

prob. that atleast one train arrives in t mins = 1 - e-(t/6)

now , 0.95 =   1 - e-(t/6)

e-(t/6) = 1 - 0.95

-(t/6) = ln(0.05)

- t/6 = - 2.995732274

t = 17.9744 mins    are required to be 95% sure that a train will arrive

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