0
$\begingroup$

So I'm trying to answer this question:

Every customer has an average of a purchase every 10 minutes. That equals 1/600 purchases per second.

My question is: How many customers do I need in order to have a 1% chance of having 30 purchases per second?

I think that I should use poisson here, but kind of like 'reverse' poisson where I know the P(X) and I need to find out the mean. After that I just divide the mean by 1/600 and I get the number of customers. Is that approach correct?

$\endgroup$
1
  • 1
    $\begingroup$ I'd say the problem is not well posed. First of all: what does average purchase mean? If a customer makes no purcheases during the year but then buys a lot of stuff on 31 dec, is it considered a purchase every 10 minutes? If so, the problem has no solution. Even if we consider that each customer has a rv $X$ which is "time till he purchases again", I doubt one can go from the mean of such a distribution to a precise probability $\endgroup$ – Ant Apr 11 '16 at 20:56
1
$\begingroup$

Hints:

For each number of customers, you'll have a different poisson distribution. The rate of that distribution will be proportional to the number of customers.

Solve the following equation for N:

$P(Poisson(N/600) > 30) > 1\%$

$\endgroup$
1
  • $\begingroup$ Yup, this worked. Thanks a lot =) $\endgroup$ – Gaspa79 Apr 12 '16 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.