Poisson Distribution

First off, how many total days are we talking here? Months have different lengths, and the bank, let's suppose, is only open on six days out of every seven. Taking an average case, we may assume 26 days as the time during which Zena places her bets with herself. Then we want to know, in 26 trials, given an average rate of r = 6, how many days will there probably be when the actual observed number is 4? There is no harm in looking at the appropriate column of the Poisson 2 table in its entirety, together with the corresponding expectations (E) for a series of 26 trials. Here, then, is the answer to the question asked, and the answer to every other similar question that might be asked of the same situation:

	r = 6	E
p(0)	0.0025	0.0650 ~ 0
p(1)	0.0149	0.3874 ~ 0
p(2)	0.0446	1.1596 ~ 1
p(3)	0.0892	2.3192 ~ 2
p(4)	0.1339	3.4814 ~ 3
p(5)	0.1606	4.1765 ~ 4
p(6)	0.1606	4.1765 ~ 4
p(7)	0.1377	3.5802 ~ 4
p(8)	0.1033	2.6858 ~ 3
p(9)	0.0688	1.7888 ~ 2
p(10)	0.0413	1.0738 ~ 1
p(11)	0.0225	0.5850 ~ 1
p(12)	0.0113	0.2938 ~ 0
p(13)	0.0052	0.1352 ~ 0
p(14)	0.0022	0.0572 ~ 0
p(15)	0.0009	0.0234 ~ 0
p(16)	0.0003	0.0078 ~ 0
p(17)	0.0001	0.0026 ~ 0

In the Expected values portion of the row for p(4), which supplies the probability of getting 4 bad checks when the average rate is 6, we find that the likely number is 3 such days (the exact answer does not quite round up to 4). So Zena will be triumphant not quite once per average week during her month. Not too bad.

Zena, who is a good manager, can learn more from the table than just that, and this might be a good moment to point out some of it. Part of it is review, part of it is new; all of it is relevant to Poisson behavior.

First, and Zena carefully checks this more or less by instinct, the days in the E column add up to 25, not 26, as was specified in the problem. This is because the probabilities for p(12) and higher do not individually register as 1 day, but collectively, they do. The missing day is in those numbers. If we had made a row for p(12-15), its aggregate likelihood would have been 0.0196, and the days of that kind to be expected in 26 trials would have been 0.5096 ~ 1. The balance of the table could then be summarized as p(16+), meaning 16 or more bad checks per day, and that row would validly have shown an E value of zero. After that adjustment, we can say that Zena can count on having less than 16 bad checks to deal with, even on the bank's worst day. Of course, the bad check writers may go on a crime spree next week, a thing which cannot be predicted from past performance, but this counts as disaster planning, not routine planning. There is a place for routine planning, and that is what we are discussing here.

Second, the likeliest number of bad checks per day is 6, as specified, but that exact number will probably occur only on 4 days. The adjacent values, for 5 or 7 bad checks per day, are just as likely: that number will also probably be experienced on 4 days each. In terms of days, then, and after rounding, the rate r = 6 actually works out to - actually means - a likeliest result which is indifferently 5, 6, or 7 checks per day.

Third, given the average rate of 6 per day, the number of bad checks to be expected on any reasonably routine day ranges from 2 to 11, and it is very unlikely that the bank will see a day with no bad checks coming in. Thus the bad check expert will need to be available pretty much all the time. (This would not be true with a smaller average rate, but 6 per day is quite a lot). That specialist will not have much to do most of the time, so that a staffer with more than one function is indicated. This is how we design the personnel profile of the bank. (For more on bank operational design, go back to Answer 10).

Fourth, the average, and single most likely, event is 6 checks, but that precise result, as noted above, will occur only on 4 days out of the 24. If we think about it too long, this raises the ultimately undecidable question, "How probable is probability?" The answer in this case is that the likeliest single answer will actually occur only 4/26 or 15.38% of the time. Everything else will be one or another of the less likely results. This may seem counterintuitive, but it is correct. The less likely results, taken together, bulk large.

Finally, given that this is a Poisson situation, a result higher than the average is a little more likely (such results, according to our table, will occur on 11 out of the 26 days) than one lower than the average (which by the same table may be looked for on 10 days). This is because the Poisson distribution is always asymmetrical, and is always skewed to the high end rather than the zero end. If empirical observation in future should show that the pattern of bad checks is perfectly symmetrical, with variation up equal to variation down, then we after all do not have a Poisson situation, and other laws of nature apply instead.