How many bags must the store have on hand at the start of each week to ensure that the probability they run out of oats is less than 0.05?A supermarket sells on average 5 10kg bags of rolled oats...
How many bags must the store have on hand at the start of each week to ensure that the probability they run out of oats is less than 0.05?
A supermarket sells on average 5 10kg bags of rolled oats per week. Assume that the number of bags sold per week is a poisson variable.
To solve this question, we need to examine the Poisson Distribution:
`P(N=k) = (lambda^k e^(-lambda))/(k!)`
Here, `lambda` is the expected number of occurrances per week, and `k` is the number of purchases for which we're checking the probability. We know, therefore, that `lambda = 5`.
Our goal here, then, is to sum the probabilities from k = 0 to k = n until the sum is above 95%, which would give us a probability that is less than 0.05 that that number of bags would not be enough to keep the sheves stocked. Start with k = 0, the case where no oats are purchased in a week:
If k = 0, `P(N = 0) = (5^0e^-5)/(0!) = e^-5 = 0.006738`
If k = 1, `P(N = 1) = (5^1 e^-5)/(1!) = 5e^-5 = 0.033690`
Our cumulative probability is now 0.040428.
Continuting in the same way, `P(N = 2) = 0.084224` giving us a cumulative probability of 0.124652.
`P(N=3) = 0.140374` giving us a cumulative probability of 0.265026.
`P(N=4) = 0.175467` giving us a cumulative probability of 0.440493.
`P(N=5) = 0.175467` with cumulative probability of 0.615961.
`P(N=6) = 0.146223` with cumulative probability of 0.762183.
`P(N=7) = 0.104445` with cumulative probability of 0.866628.
`P(N=8) = 0.065278` with cumulative probability of 0.931906.
`P(N=9) = 0.036266` with cumulative probability of 0.968172.
Therefore, we have a cumulative probability of roughly 96.8% that 9 bags or less will be purchased. Therefore, to never run out of oats, with a probability of 0.95 we would need at least 10 bags at the beginning of the week.
Notice that I also included the probability that no bags are purchased (P(N=0)). This is important in these situations so that our number is not thrown off by not including a very real possibility!
Unfortunately, as far as I know, there is no elegant way to do this without computers doing the above for you. In problems like this, you just need to find the probabilities and add as you go!
I hope that helps!