1. ## Poisson distribution

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 $\lambda =0.4$. 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.

2. The Poisson Distribution is perfectly scalable.

1 guy -- $\lambda_{1} = 0.40$

2 folks -- $\lambda_{2} = 2*0.40 = 0.80$

35 homies -- $\lambda_{35} = 35*0.40 = 14.00$

9432 peeps -- $\lambda_{9432} = 9432*0.40 = 3772.80$

Where does that leave us?

3. Originally Posted by TKHunny
The Poisson Distribution is perfectly scalable.

1 guy -- $\lambda_{1} = 0.40$

2 folks -- $\lambda_{2} = 2*0.40 = 0.80$

35 homies -- $\lambda_{35} = 35*0.40 = 14.00$

9432 peeps -- $\lambda_{9432} = 9432*0.40 = 3772.80$

Where does that leave us?

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.

4. 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?

5. I'm still not quite sure what to do, so I should try to use Normal Approximation. But, I would be able to calculate the probability that enough beers, so the should the probability of not enough beers equal to 1 minus the probability of enough beers?

6. ??? 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%?

7. I tried to do something like this:

$P(Not\ enough\ beers)=1-P(Enough\ beers)<0.05$
$1-P(z\leq \frac{x-8000}{40\sqrt{5}})<0.05$

Then I solved for x and I got the answer is $x=8147$, 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.

8. What is $40\cdot\sqrt{5}$?

It's Poisson! mean = standard deviation = 20000*0.40

9. Originally Posted by TKHunny
What is $40\cdot\sqrt{5}$?

It's Poisson! mean = standard deviation = 20000*0.40
Sorry I'm confused, because I saw my notes say for Poisson mean=variance=parameter, which is 20000*0.4, that's why I squared it to get standard deviation.

10. It appears I need a nap. You have it right. Many apologies.