statistics
posted by maria .
A service station has a pump that distributes diesel fuel to automobiles. The station owner estimates that only about 3.2 cars use the diesel pump every 2 hours. Assume the arrivals of diesel pump users are Poisson distributed.
a. What is the probability that three cars will arrive to use the diesel pump during a 1hour period?
b. Suppose the owner needs to shut down the diesel pump for half an hour to make repairs. However, the owner hates to lose any business. What is the probability that no cars will arrive to use the diesel pump during a halfhour period?
c. Suppose five cars arrive during a 1hour period to use the diesel pump. What is the probability of five or more cars arriving during a 1hour period to use the diesel pump?

Poisson distribution (m = mean):
P(x) = e^(m) m^x / x!
Determining m:
3.2 cars every 2 hours
1/2(3.2) cars every hour = 1.6
1/4(3.2) cars every 1/2 hour = 0.8
For a):
x = 3
m = 1.6
Substitute into the formula and find the probability.
For b):
x = 0
m = 0.8
Substitute into the formula and find the probability.
For c):
x = 0,1,2,3,4
m = 1.6
Find the probabilities and add for a total. Take that value and subtract from 1.
I hope this will help get you started.