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You are tasked with ordering bottled water for an upcoming outdoor event. Bottles cost you $0.10 a piece and you can sell them that night for $2....

Since we make them for $0.10 and sell them for $2.00, each bottle we sell will make us a profit of $1.90.

But since we can only resell them later for $0.05, each bottle we make but do not sell will force us to take a loss of $0.05.

We don't know how many we will sell, but we know that it is normally distributed with a mean of 10,000 and a standard deviation of 1,000. The goal is to find the amount of production that will maximize our expected profit.


To do this, we want to find the point where the marginal expected profit from making another bottle is zero, which is the point where the probability of selling that bottle and making $1.90 is exactly balanced by the probability of not selling that bottle and losing $0.05.
Thus, we know that the following holds, at the desired number of bottles n, with p(n) being the probability of selling the nth bottle:


`p(n) (1.90) - (1-p(n)) (0.05) = 0`


`1.90 p(n) + 0.05 p(n) - 0.05 = 0`


`1.95 p(n) = 0.05`
`p(n) = 0.05/1.95 = 0.0256`


This is essentially a p-value; it's the probability of selling at least that many bottles, when the number of bottles sold is normally distributed. Only the right-tail p-value is important here.

This is actually extremely close to the p-value of z=2, 2 standard deviations above the mean. A more precise z-score is actually 1.95; we want to make enough bottles to cover 1.95 standard deviations above the mean. Below that point, we're likely enough to sell those bottles that it's worth the risk; above that point, the probability of selling them is too low to be worth taking the risk.

Since the mean is 10,000 and the standard deviation is 1,000, 1.95 standard deviations above the mean is 11,950 bottles--just shy of 12,000.

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