# The Poisson Distribution?

• Mar 27th 2010, 10:42 AM
mintsharpie
The Poisson Distribution?
Customer arrivals at a checkout counter have a Poisson distribution with an average of 7 per hour. For a given hour, find the probabilities of the following events:

a) Exactly seven customers arrive
b) No more than two customers arrive
c) At least two customers arrive

Can someone help me figure this out? I don't know how to approach this problem.
• Mar 27th 2010, 11:15 AM
TKHunny
These can be calculated directly.

$p(7) = \frac{e^{-7}\cdot 7^{7}}{7!}$

p(No more than 2) = p(0)+p(1)+p(2)

p(at least 2) = 1- (p(0)+p(1))

Let's see what you get.
• Mar 27th 2010, 12:13 PM
mintsharpie
Quote:

Originally Posted by TKHunny
These can be calculated directly.

$p(7) = \frac{e^{-7}\cdot 7^{7}}{7!}$

p(No more than 2) = p(0)+p(1)+p(2)

p(at least 2) = 1- (p(0)+p(1))

Let's see what you get.