The Poisson Distribution is perfectly scalable.
1 guy --
2 folks --
35 homies --
9432 peeps --
Where does that leave us?
Hi, I got this question here and I found it is really tricky to solve.
20000 people are watching football match at the stadium. The number of cans of beer that a spectator wants to drink is distributed according to a Poisson distribution with parameter . How many cans of beer do organizer need to stock so that the probability that there is not enough beer is less than 0.05?
I have no idea how to start, can anyone please give me some advice? Thanks a lot.
Thanks a lot, I got what you mean. But I still don't know how to solve the question. So, is this the kind of situation that Poisson distribution with "large" parameter, so should I use Normal approximation? The one that use Normal distribution to solve the problem? Please kindly give me more advice. Thanks a lot.
You could use a normal approximation with mean = standard deviation = 20000*0.40. In fact, due to the difficulty in calculating directly such Poisson probabilities, I would recommend the Normal Approximation. Just leave 0.05 in the right tail.
Why is it difficult to calculate directly the Poisson probabilities?
??? p(too many customers) = 0.05. This is a z-score lookup.
Using the empirical rule:
p(customers exceeds the mean) = 50%
p(customers exceed mean + 1 standard deviation) = 84%
p(customers exceed mean + 2 standard deviation) = 97.5%
What will it take to get 99.5%?
I tried to do something like this:
Then I solved for x and I got the answer is , by using table of standard normal for z values.
So that means there are 8147 cans of beer need to be stocked.
Did I do right things? I just kind of feel the number 8147 is not reasonable enough.