STAT-UB.0103 NOTES for Wednesday 2012.FEB.15 Suppose that a hospital has a cardiac care unit which handles heart attac victims on the first day of their problems. The geographic area served by the hospital creates new patients as observations on a Poisson phenomenon, with a mean rate of 4.1 per day. If the hospital has 5 beds, what is the probability that they will have an overload problem? Now find the minimum number of beds so that the overload probability is reduced to 5% or less. In this context, the thing we call the decision variable is K, the number of beds. Our solution will be done by stating a value for K. The objective here is to minimize K; that is min K is what we are trying to do. We would say that K is the objective function. In this simple problem, K is both the objective function and the decision variable. In more complicated problems, the objective function might be 45 Q - 10 R. We will need to have P[ X K ] 0.95. This statement constitutes a constraint. Problems can have many constraints. Let X be the random number of arrivals on any days. Let K, the number of beds, be the decision variable. Value Prob Cumulative 0 0.016573 0.01657 1 0.067948 0.08452 2 0.139293 0.22381 3 0.190368 0.41418 4 0.195127 0.60931 5 0.160004 0.76931 6 0.109336 0.87865 7 0.064040 0.94269 8 0.032820 0.97551 9 0.014951 0.99046 10 0.006130 0.99659 11 0.002285 0.99887 12 0.000781 0.99966 13 0.000246 0.99990 14 0.000072 0.99997 15 0.000020 0.99999 16 0.000005 1.00000 1
We can see that With 5 beds, P[adequate] = P[X 5] = 0.76931 77% With 6 beds, P[adequate] = P[X 6] = 0.87865 88% With 7 beds, P[adequate] = P[X 7] = 0.94269 94% With 8 beds, P[adequate] = P[X 8] = 0.97551 98% It loos lie they re going to need 8 beds. With a mean of 4.1, the problem is clearly that 8 beds will given them an expected vacancy rate of about 8 4.1 = 3.9 beds per day. This is 8 4.1 8 49% of their capacity. = 3.9 8 This example can be done in many different contexts. Suppose that you were running a canoe rental shop in which people come in for canoes early in the day and that the canoes will be returned in late afternoon. How many canoes should you have? It s the same logic. ( X ) ( X ) SD λ 1 For the Poisson random variable X, the ratio is =. This will E λ λ be treacherous to manage when λ is small. For our example with λ = 4.1, this is 1 0.49. That is, the standard deviation is about half the mean. Thus very 4.1 low values (near zero) are easily possible, and this does awful things to the utilization rate. Fine. So what should be done about this? The only hope is to redesign the situation to deal with Poisson random variables with larger means. For the hospital, their liely solution will come in mergers (at least for cardiac care) with nearby institutions. In a consortium of hospitals, each will have a different speciality. There are several interesting ways in which the Poisson probability function can be obtained as a mathematical exercise. (These are not official course content, and they will not be covered on homewors or exams. It s OK to sip ahead to page 4.) (1) Consider the random variable X which denotes the number of events on the time interval (0, t). Divide this up into n subintervals, each of length t n. Assume 2
that the occurrence of events in different subintervals is statistically independent. Assume also that the subintervals are so small that the probability of two or more events is essentially zero. If p represents the probability of an event in any interval, then the total number of events, X, is binomial, so that n p p x P[ X = ] = ( 1 ) n Now let n go to infinity. As this happens, the subintervals will get shorter and shorter, so the value of p will more as well. Assume that np goes to λ as n goes to infinity. You can show that n lim p 1 p n x np λ ( ) n = e λ λ! (2) Consider the random variable X which denotes the number of events on the time interval (0, t). Let f (t) be the probability that there are events in (0, t) ; that is, f (t) = P[ X = ]. Now consider a very short interval of length dt. Assume that P[ no events in the interval of length dt ] 1 λ dt P[ one event in the interval of length dt ] λ dt P[ two or more events in the interval of length dt ] 0 If you see a formal mathematical description of this, you ll find that the right sides above are, respectively, 1 λ dt + o(dt) λ dt + o(dt) o(dt) The o(dt) symbols are used to denote something that goes to zero even faster than dt. The formal notion is o( dt) lim = 0. Examples of things that would qualify as dt 0 dt o(dt) are (dt) 2 and (dt) 3. These assumptions assume that the probability of an event in a short interval is approximately proportional to the length of the interval. These also assume that the probability of two or more events in a short interval is essentially zero. Then f ( t + dt) f ( t ) [ 1 λ dt ] + f -1 ( t ) [ λ dt ] 3
In probability symbols, P[ events in (0, t + dt) ] P[ events in (0, t) ] P[ no events in (t, t + dt) ] + P[ -1 events in (0, t) ] P[ one event in (t, t + dt) ] Next write f ( t + dt) f ( t ) + f ( ) t dt. This is an instance of Taylor s theorem from calculus. This gets us to f ( t ) + f () t dt f ( t ) [ 1 λ dt ] + f -1 ( t ) [ λ dt ] The simplification is f () t dt λ f ( t ) dt + f -1 ( t ) [ λ dt ] and, after cancelling the dt, f () t λ f ( t ) + λ f -1 ( t ) This system of equations, starting with = 1, will lead to the Poisson in the form f (t) = e λt ( λt)! (This ends the discussion on extra material.) Finally, let s note the hypergeometric random variable. This is illustrated by the following basic quality control situation. Suppose that a shipment of 60 garments has been received, and it is to be inspected for quality. The inspection consists of a random selection of 6 garments. Consider, hypothetically, that there are 5 defectives among the 60. Find the probability that exactly 2 defectives turn up in the sample. 4
5 There are = 10 ways to select 2 of the 5 defectives to turn up in the sample. 2 However, we also will be selecting 4 non-defectives to fill up this sample, and that can be 55 5 55 done in ways. Thus, there are = 10 341,055 = 3,410,550 ways to 4 2 4 obtain a sample with 2 defectives and 4 non-defectives. Of course, the total number of 60 ways to select 6 things from 60 is = 50,063,860. Thus, the probability of getting 6 3,410,550 exactly 2 defectives in the sample is 50,063,860 0.068124. In general notation, let N be the number of items in the whole lot (or population) let M be the number of defectives in the whole lot let n be the size of the sample Then the probability of getting exactly x defectives in the sample is M N M x n x N n If we use random variable X to denote the number of defectives in the sample, then we would write P[ X = x ] = M N M x n x N n 5
Minitab can be used for this difficult calculation. Use Calc Probability Distributions Hypergeometric and fill in the panel as follows: Minitab calls M the event count in population. The event could be a success, it could be a defect, but in general it s whatever we re counting. Probability Density Function Hypergeometric with N = 60, M = 5, and n = 6 x P( X = x ) 2 0.0681240 In fact, the common card problems are related to the hypergeometric distribution. Suppose that you deal out 13 cards, as in the game of bridge. What is the probability that you will have two aces? This is a hypergeometric situation with N = 52, M = 4, n = 13. 4 48 2 11 The probability that you get exactly x = 2 aces is 0.2135. The Minitab 52 13 wor is shown below: Probability Density Function Hypergeometric with N = 52, M = 4, and n = 13 x P( X = x ) 2 0.213493 6
In fact, you can get the whole distribution for the number of aces: Aces Probability 0 0.303818 1 0.438848 2 0.213493 3 0.041200 4 0.002641 Here is the whole distribution for the number of any particular suit, say : Probability 0 0.012791 1 0.080062 2 0.205873 3 0.286330 4 0.238608 5 0.124692 6 0.041564 7 0.008817 8 0.001167 9 0.000093 10 0.000004 11 0.000000 12 0.000000 13 0.000000 For business purposes, of course, the major use of the hypergeometric is in quality control inspections. For example, suppose that a shipment of 100 cell phones has exactly 8 defective. Suppose that you tae a random sample of 5 to chec up on these. The probability that your sample will have one defective (and thus 4 non-defectives) is F8 92 HG I 1K J F H G I 4 K J F100I HG 5 K J = 8 2,794,155 75,287,520 0.2969 The probability, for comparison, of having no defectives in the sample is F8 92 HG I 0K J F H G I 5 K J F100I HG 5 K J = 1 49,177,128 75,287,520 0.6532 7
The 8% defective rate is large enough to be annoying. Notice, though, that your random sample of 5 has a 65% chance of looing perfect. Suppose that we each have a dec of cards. You select 8 cards at random and I select 5 cards at random. What is the probability that we have exactly 2 matches? (A match here would mean that we both select, say, 4.) Let s suppose that you select first. In essence, you ve divided the dec into 8 hot cards and 44 cold cards. Now, if my selection is random, the probability that I get 2 hot and 3 cold is 8 44 2 3 = 52 5 28 13,244 2,598,960 0.1427 This can be conceptualized in the other order. Suppose that I select first. I ve divided the dec into 5 hot and 47 cold cards. Now when you select your 8 cards, the probability that you get 2 hot and 6 cold is 5 47 2 6 52 8 = 10 10,737,573 752,538,150 0.1427 You can see that this is just a variation on the QuicDraw game, or on Lotto or other lottery games. OK, let s wor out that example. In QuicDraw, there are 80 numbers, and the computer selects 20 of them. There are choices as to how many of them I select. Suppose that I choose to select 8. Let s find the probabilities for 8 winners, 7 winners, The action of the QuicDraw computer creates 20 hot numbers and 60 cold numbers. NOTE: We could thin also of my selection maing 8 hot and 72 cold numbers and then asing how many the QuicDraw computer would hit. 8
Now if I select 8 numbers, the probability that I get hot and 8- cold is 20 60 8 80 8 This is a pain to compute. Since we ll need 80 8 several times, let s compute 80 8 80 79 78 77 76 75 74 73 = 8 7 6 5 4 3 2 1 = 28,987,537,150 For the case = 8, all winners, the numerator is 8 0 probability) is 0.4346 10-5. Not a good chance. = 125,970 and the ratio (the Now let s try = 7. Still a good game. The numerator is 7 1 the ratio is 0.1605 10-3 = 0.0001605. Not liely either. = 4,651,200 and For = 6, the numerator is = 68,605,200 and the ratio (the probability) is 6 2 0.002367. This is a tiny number, but at least we can thin about it without scientific notation. For = 5, the numerator is This is just under 2%. For = 4, the numerator is just over 8%. 5 3 4 4 = 530,546,880. The probability is 0.01830. = 2,362,591,575. The probability is 0.08150, 9
For = 3, the numerator is matches is then 0.2148. 3 5 = 6,226,123,680. The probability of getting 3 You can see that these games frequently result in uninteresting outcomes. The probability of two matches is 0.3281. The probability of one match is 0.2665. Actually, there s a sort of booby prize if you get zero matches. This occurs with probability 0.0883. Here is a summary: Number Probability of winners 0 0.0883 1 0.2665 2 0.3281 3 0.2148 4 0.08150 5 0.01830 6 0.002367 7 0.0001605 8 0.000004346 Total 1.000031846 You can get Minitab to produce this list. Here s how. Begin by putting the numbers 0 through 8 down C1. (You can use Calc Mae Patterned Data Simple Set of Numbers.) Do Calc Probability Distributions Hypergeometric. In the resulting panel, push the radio button for Probability. Then, in the middle set of boxes, Set Population (N): to 80 Set Successes in Population (M): to 20 Set Sample size (n): to 8 Set Input column: to C1 Set Optional storage: to C2 10
If you want to see more precision in C2, highlight C2 and then do Editor Format column Numeric. You can as for Fixed and as for, say, 12 figures after the decimal. In terms of the symbols N (lot size), M (number of special items), n (sample size), and X (random number of special items in sample), we have The mean of X, written E(X), is M n N. M M N n The standard deviation of X, written SD(X), is n 1. N N N 1 Usually the symbol M corresponds to what we do not now. We assume hypothetical values for M to relate to the actual values of X which we see. 11