# Poisson distribution

• Apr 29th 2011, 07:13 PM
tsang
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.
• Apr 29th 2011, 10:01 PM
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?
• Apr 30th 2011, 06:30 AM
tsang
Quote:

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.
• Apr 30th 2011, 07:22 PM
TKHunny
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?
• Apr 30th 2011, 07:37 PM
tsang
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?
• Apr 30th 2011, 08:06 PM
TKHunny
??? 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%?
• Apr 30th 2011, 08:08 PM
tsang
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.
• Apr 30th 2011, 08:18 PM
TKHunny
What is $40\cdot\sqrt{5}$?

It's Poisson! mean = standard deviation = 20000*0.40
• Apr 30th 2011, 08:26 PM
tsang
Quote:

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.
• Apr 30th 2011, 09:22 PM
TKHunny
It appears I need a nap. You have it right. Many apologies.