# Thread: probability distribution function question

1. ## probability distribution function question

Here is my dilemma:

A chain of stores is concerned about the number of complaints in regards to customer service. Historical data shows that the fraction of complaints is around .0040 a day. On one day it is estimated that about 5000 transactions occur throughout the chain (each transaction is subject to one complaint).

My first question,
1) Is this a poisson PDF?
2) What is the probability that there are no complaints?
3) How many complaints should we expect in one day?
4) Finally, what are the chances we receive more than 25 complaints today?

Any help would be greatly appreciated.

2. Originally Posted by Stormy-1
Here is my dilemma:

A chain of stores is concerned about the number of complaints in regards to customer service. Historical data shows that the fraction of complaints is around .0040 a day. On one day it is estimated that about 5000 transactions occur throughout the chain (each transaction is subject to one complaint).

My first question,
1) Is this a poisson PDF?
2) What is the probability that there are no complaints?
3) How many complaints should we expect in one day?
4) Finally, what are the chances we receive more than 25 complaints today?

Any help would be greatly appreciated.
It's a binomial distribution.

0.004 / 5000 = 0.0000008 complaints per transaction

B(5000, 0.0000008)

Assuming there are about 5000 transactions EVERY day.

HOWEVER...

As p is large and n is small, a poisson approximation Po(np) would be suitable in this case...

i.e. Po(0.04)

Probability of no complaints is (0.04^0)(e ^ -0.04) / 0! = 0.961

Expected number of complaints is 0.04 as that's the average.

p(x > 25) = 1 - p (x <= 25) : Use the poisson table.

3. Originally Posted by sean.1986
It's a binomial distribution.

[deleted]

B(5000, 0.004)

[deleted]

HOWEVER...

As p is small and n is large, a poisson approximation Po(np) would be suitable in this case...

i.e. Po(0.04)

Probability of no complaints is (20^0)(e ^ -20) / 0! ~ 0.

Expected number of complaints is 20 (see below)

p(x > 25) = 1 - p (x <= 25) : Use the poisson table.
A couple of corrections (in red).

Note: The mean for the Poisson random variable X (the number of complaints) in this situation is $\mu = (0.004)(5000) = 20$.

4. ## Thanks

Thanks for your help, Greatly appreciated.