Which distribution do I need to use?

Question

In a shop, the average customers per 5 minutes is 3. What is the probability that the shopkeeper has to wait at least 6 minutes before the second customer walks in.

I don't know which distribution I have to use and how to solve this question. I was thinking about Poisson, negative binomial, exponential maybe, but I honestly don't know how to solve this.

J.G. · Answer 1 · 2018-12-20T23:19:54.267

0

The number of customers per unit time is Poisson. Equivalently, the time between consecutive customers being exponentially distributed, the two variables' means having product $1$. The sum of two exponential iids has a $k=2$ Erlang distribution. I'll leave the rest to you.

edited Dec 20 '18 at 23:19

answered Dec 20 '18 at 23:13

J.G.

115,835

Thanks! But the only thing is that we didn't study the Erlang distribution in class. Only Poisson and exponential. – Dries De Witte Dec 20 '18 at 23:23
@Dries Dr Witte Fair enough. Have you studied the Gamma distribution? It's a special case of that. Failing that, try to derive the distribution of the sum of two exponential iids. You'll find using either moment generating functions or characteristic functions helps, if you've studied them. – J.G. Dec 20 '18 at 23:25
No, we didn't study all those things. Just Poisson, negative binomial, exponential. – Dries De Witte Dec 20 '18 at 23:36
@DriesDeWitte You might need to use convolution instead then, but they've certainly set you a hard question, unless they meant the time between the first two customers. Then you just need the exponential distribution itself. – J.G. Dec 20 '18 at 23:38
I think they mean the time between the first two customers indeed. Can you help me find the values of the parameters for this exponential distribution? – Dries De Witte Dec 21 '18 at 00:01
@J.G. is correct: Poisson. – David G. Stork Dec 21 '18 at 00:59

score 0 · Accepted Answer · answered Dec 21 '18 at 02:10

If the question means, right now, before any customers have walked in, what is the probability that we will need to wait at least 6 minutes before the second customer walks in? then we might interpret it as what is the probability that either no customers or only one customer arrives in the next 6 minutes (in which case the second customer could only arrive after 6 minutes)?

In that case we could remodel the situation as $X\sim Po(\lambda_6)$ where the $\lambda_6=\frac{18}{5}$ is the average number of customers per 6 minutes (based on 3 customers per 5 minute period). Then, $P(X=0\lor X=1)\approx 0.126$.

Which distribution do I need to use?

2 Answers2