# Homework Help: Poisson DIstribution

1. Dec 26, 2011

### jsmith613

1. The problem statement, all variables and given/known data
A hardwhere store sells on average 8 drills per week.
The store recieves ONE delivery of drills at the same time each week.
Find the no. of drills that need to be in stock after a delivery for there to be at most a 5% chance of the store NOT having sufficent drills to meet demand before the next delivery.

2. Relevant equations
3. The attempt at a solution

Ok so we know that Y≈Po(8)
where Y = no. expected to sell
lets say n = demand no.

thus we want
p(n > Y) < 0.05
so
P(Y<n) < 0.05

P(Y>n) < 0.05

How is this possible
thanks

2. Dec 26, 2011

### micromass

Aren't your Y and n thesame thing. The number expected to sell surely equals the demand??

I think your n should be something like "drills in store".

3. Dec 26, 2011

### jsmith613

how can no. expected = demand

the Poisson value = E(X) but demand could be greater than or smaller than this, surely

4. Dec 26, 2011

### jsmith613

ahh
Y is a distribution which means it could take on OTHER values but not def.
so Y = demand
n = no. in stock
with Y-P(8)

So
P(Y>n) <= 0.05

I see
ok thanks

5. Dec 26, 2011

### Ray Vickson

You need to distinguish between sales and demand. If demand = X~Po(8) and we start the week with stock N, then sales S = min(X,N); that is, sales = demand if demand <= N, and sales = N otherwise. You need to choose N so that P(X > N) <= 0.05; that is, there is a chance of no more than 5% that demand exceeds on-hand stock. (By the way, that is <= 0.05, not < 0.05, because that is what 'no more than' means.)

RGV

6. Dec 27, 2011

cheers :)