# Poisson distrubution

• Dec 22nd 2009, 01:55 AM
Allan89a
Poisson distrubution
Nr of people entering a store per min. is 3.8.

If 30 non-overlapping minutes are picked, then what is the probability that in at least 20 of these minutes, at least 4 people arrived to the store.

I've calculated for P(X>=4) , which was approximately 0.5265.

But how do you go on from here. Is it binomial now? Should I now approximate or use the central limit theorem or something?

Btw, answer to question is 0.09, but how to solve?
• Dec 22nd 2009, 02:04 AM
mr fantastic
Quote:

Originally Posted by Allan89a
Nr of people entering a store per min. is 3.8.

If 30 non-overlapping minutes are picked, then what is the probability that in at least 20 of these minutes, at least 4 people arrived to the store.

I've calculated for P(X>=4) , which was approximately 0.5265.

But how do you go on from here. Is it binomial now? Should I now approximate or use the central limit theorem or something?

Btw, answer to question is 0.09, but how to solve?

Let Y be the random variable 'number of non-overlapping minutes in which at least 4 people arrived'.

Y ~ Binomial(n = 30, p = 0.5265).

Calculate \$\displaystyle \Pr(Y \geq 20)\$.

I get 0.0868.
• Dec 22nd 2009, 06:47 AM
Allan89a
Quote:

Originally Posted by mr fantastic
Let Y be the random variable 'number of non-overlapping minutes in which at least 4 people arrived'.

Y ~ Binomial(n = 30, p = 0.5265).

Calculate \$\displaystyle \Pr(Y \geq 20)\$.

I get 0.0868.

Hi,

Thanks for your post! I seem to have mixed up some concepts, but now I finally got to 1 -F(1.35)= 0.0885.

And yeah, the answer in the book's probably approximated to two decimals.