Poisson Distribution
Answer 15

9 jars.

Explanation

It is given that r = 5. On current assumptions, we wish to have jars left over at the end of the week (that is, no disappointed customers) not less than 90% of the time. How do we start to work out the answer?

If you can't see how to start, start high and work by trial and error. The highest value for which the r = 5 column of the Poisson Table finds it worthwhile to report a value is p(15) = 0.0002. This means that it is very unlikely that Filbert will sell 15 jars a week, so that buying 15 jars would give him an overwhelmingly strong guarantee that no Saturday customers will be disappointed.

But Filbert cannot afford to give that strong a guarantee. It is also unlikely that he will sell 14 jars, since from the table p(14) = 0.0005. The chance of his selling 14 or 15 jars is their sum, 0.0007. That contingency is still remote, and providing for it is very expensive. It is much more protection (99.93%) than Filbert bargained for. If we continue to add probabilities, moving upward on the table, we will sooner or later reach a cumulative result which will exceed the 10% that Filbert accepted as his risk of disappointed customers. That point is reached at p(8), when the cumulative probability has become 0.1333, or more than 13%. This means that there is a better than 13% chance that Filbert, if he had them, would sell 8 or more jars of frog legs. At that purchase level, he is below his agreed threshold, and to avoid that high a disappointment rate, we must back up one grade on the table.

The optimum purchase is thus 9 jars, and the chance of disappointing further frog leg customers at that rate is 0.0680. That is, the lowest acceptable guarantee level turns out to be not exactly 90%, but 93%. Only 7% of the time (more precisely 6.8% of the time) will discerning supermarket customers be turned away froglegless.

Presumably Filbert's store will now attract all the discerning diners in the region, and become immensely profitable??

Comment

We may note that a variant of this problem was solved in Fortune Magazine, issue of 7 Mar 1994. They also reached the answer of 9 jars. Could anything be more secure?

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