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**A busy railway station has two taxi stands at separate exits, A. At stand A, taxis arrive according to a Poisson Process of rate 2 per minute.

Following the arrival of a train, a queue of 40 customers suddenly builds up at stand A, waiting for taxis to arrive. Assume that whenever a taxi arrives, it instantly picks up exactly one customer (if there is one), and assume that all customers in the queue patiently wait their turn. Approximately calculate the probability that the 40th customer in the queue has to wait longer than 15 minutes in total before boarding a taxi.**


Not quite sure which distribution this question is hinting at. My first thought was that it was the normal distribution but the question doesn't hint at a standard deviation, so that I can put into the equation.

Second thought was that it could be the poison distribution but that's about finding the probability of the number of times an event happens in a fixed time interval.

thoughts?

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    $\begingroup$ I dont get how it matters whether or not there is an exit B $\endgroup$ – glowstonetrees Apr 18 '18 at 19:58
  • $\begingroup$ @glowstonetrees oh lemme edit that part out. That's for another part to the question. $\endgroup$ – blaaaaaaa Apr 18 '18 at 20:01
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By definition, something has a "Poisson process of rate $\lambda$" if $N_t\sim \text{Poisson}(\lambda )$ for all $t>0$, where $N_t$ is the number of arrivals within time $t$. Note that it has the memoryless property, so the number of arrivals in any two disjoint time sets are independent.

So in this case, having a "Poisson process of rate $2$ per minute" means that the number of taxis arriving at any given minute follows a Poisson$(2)$ distribution.

We are interested in the number of arrivals in the first $15$ minutes. Suppose that the number of taxis in the first $15$ minutes is $X$. Then $X \sim \text{Poisson}(30)$ (since the mean for one minute is $2$, the mean for $15$ minutes would be $30$).

The question is basically asking you the value of $\Bbb P(X<40)$.

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  • $\begingroup$ so is there a faster way to find P(x < 40) ? instead of doing P(x = 0) + P(x = 1) ... + P(x = 39) ? $\endgroup$ – blaaaaaaa Apr 18 '18 at 21:11
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    $\begingroup$ No, I am afraid not. You will have to use a calculator for this. $\endgroup$ – glowstonetrees Apr 18 '18 at 22:07
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Guide:

Taxis arrive according to a Poisson Process, hence you should consider Poisson distribution with mean $15\lambda$ since we should be interested in how many taxis arrive in $15$ minutes.

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