Poisson Distribution
Answer 13Probably 2.
ExplanationNothing is perfect, but there is a substantial dropoff in the Poisson predictions for the various numbers of people who do not show up for their reserved seat on our spacious and maintenance-efficient Oahu Airlines DC-95, with its 100-passenger capacity, so convenient for statistics problems.
The general rate r is 3 per 100, but as with all whole-number r, the chance of 2 no-shows is the same: both values are 0.2240. Next most frequent are 1 and 4 no-shows: 15% and 17%, respectively. Here is the whole range of possibilities, as copied from the Poisson Table:
p(0) p(1) p(2) p(3) p(4) p(5) p(6) p(7) p(8) p(9) p(10) p(11) p(12) If we count on 3 no-shows, and accordingly sell 3 extra tickets, we will as often as not find that we have 2 no-shows. This will often produce tensions between the airline and its public, and it is bad public relations. We will therefore probably avoid it. If instead we sell 2 extra tickets, we will only be embarrassed if there are 0 or 1 no-show passengers; the combined chances of this option are 0.0498 + 0.1494 = 0.1952, or roughly 1 chance in 5 (20%). How acceptable that is will depend partly on how much incentive you are willing to offer to get a passenger to release his seat, and how amusing the staff are prepared to be in making that offer. We are thus in the gray area of airline decision-making. Finally, selling only 1 extra ticket will lead to problems, on average, only 5% of the time (1 chance in 20), which is much more comfortable odds.
But comfort does not make money, at least in the eyes of those who are counting the money. The airline will probably prefer to take its chances in the gray area, and overbook by 2. At any rate, we have conjectured that strategy as our proposed answer.
Comment
Actually, the MD-95 (or, after the Boeing takeover, the Boeing 717) has a capacity of 106 passengers, but we are simplifying for pedagogical purposes.
24 Aug 2007 / Contact The Project / Exit to Resources Page
