Suppose a cereal manufacturer puts pictures of famous athletes on cards in boxes of

Transcription

1 CHAPTER 17 Probability Models Sppose a cereal manfactrer pts pictres of famos athletes on cards in boxes of cereal, in the hope of increasing sales. The manfactrer annonces that 20% of the boxes contain a pictre of Tiger Woods, 30% a pictre of David Beckham, and the rest a pictre of Serena Williams. Sond familiar? In Chapter 11 we simlated to find the nmber of boxes we d need to open to get one of each card. That s a fairly complex qestion and one well sited for simlation. Bt many important qestions can be answered more directly by sing simple probability models. 388 Searching for Tiger Yo re a hge Tiger Woods fan. Yo don t care abot completing the whole sports card collection, bt yo ve jst got to have the Tiger Woods pictre. How many boxes do yo expect yo ll have to open before yo find him? This isn t the same qestion that we asked before, bt this sitation is simple enogh for a probability model. We ll keep the assmption that pictres are distribted at random and we ll trst the manfactrer s claim that 20% of the cards are Tiger. So, when yo open the box, the probability that yo scceed in finding Tiger is Now we ll call the act of opening each box a trial, and note that: There are only two possible otcomes (called sccess and failre) on each trial. Either yo get Tiger s pictre (sccess), or yo don t (failre). In advance, the probability of sccess, denoted p, is the same on every trial. Here p = 0.20 for each box. As we proceed, the trials are independent. Finding Tiger in the first box does not change what might happen when yo reach for the next box. Sitations like this occr often, and are called Bernolli trials. Common examples of Bernolli trials inclde tossing a coin, looking for defective prodcts rolling off an assembly line, or even shooting free throws in a basketball game. Jst as we fond eqally likely random digits to be the bilding blocks for or simlation, we can se Bernolli trials to bild a wide variety of sefl probability models.

2 The Geometric Model 389 Calvin and Hobbes 1993 Watterson. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved. Daniel Bernolli ( ) was the nephew of Jacob, whom yo saw in Chapter 14. He was the first to work ot the mathematics for what we now call Bernolli trials. Activity: Bernolli Trials. Gess what! We ve been generating Bernolli trials all along. Look at the Random Simlation Tool in a new way. Back to Tiger. We want to know how many boxes we ll need to open to find his card. Let s call this random variable Y = # boxes, and bild a probability model for it. What s the probability yo find his pictre in the first box of cereal? It s 20%, of corse. We cold write P1Y = 12 = How abot the probability that yo don t find Tiger ntil the second box? Well, that means yo fail on the first trial and then scceed on the second. With the probability of sccess 20%, the probability of failre, denoted q, is = 80%. Since the trials are independent, the probability of getting yor first sccess on the second trial is P1Y = 22 = = Of corse, yo cold have a rn of bad lck. Maybe yo won t find Tiger ntil the fifth box of cereal. What are the chances of that? Yo d have to fail 4 straight times and then scceed, so P1Y = 52 = = How many boxes might yo expect to have to open? We cold reason that since Tiger s pictre is in 20% of the boxes, or 1 in 5, we expect to find his pictre, on average, in the fifth box; that is, E1Y2 = = 5 boxes. That s correct, bt not easy to prove. The Geometric Model Geometric probabilities. See what happens to a geometric model as yo change the probability of sccess. We want to model how long it will take to achieve the first sccess in a series of Bernolli trials. The model that tells s this probability is called the Geometric probability model. Geometric models are completely specified by one parameter, p, the probability of sccess, and are denoted Geom(p). Since achieving the first sccess on trial nmber x reqires first experiencing x - 1 failres, the probabilities are easily expressed by a formla. NOTATION ALERT: Now we have two more reserved letters. Whenever we deal with Bernolli trials, p represents the probability of sccess, and q the probability of failre. (Of corse, q = 1 - p.) GEOMETRIC PROBABILITY MODEL FOR BERNOULLI TRIALS: Geom(p) p = probability of sccess (and q = 1 - p = probability of failre) X = nmber of trials ntil the first sccess occrs P1X = x2 = q x-1 p Expected vale: E1X2 = m = 1 p Standard deviation: s = A q p 2

3 390 CHAPTER 17 Probability Models FOR EXAMPLE Spam and the Geometric model Postini is a global company specializing in commnications secrity. The company monitors over 1 billion Internet messages per day and recently reported that 91% of s are spam! Let s assme that yor is typical 91% spam. We ll also assme yo aren t sing a spam filter, so every message gets dmped in yor inbox. And, since spam comes from many different sorces, we ll consider yor messages to be independent. Qestions: Overnight yor inbox collects . When yo first check yor in the morning, abot how many spam s shold yo expect to have to wade throgh and discard before yo find a real message? What s the probability that the 4th message in yor inbox is the first one that isn t spam? There are two otcomes: a real message (sccess) and spam (failre). Since 91% of s are spam, the probability of sccess p = = Let X = the nmber of s I ll check ntil I find a real message. I assme that the messages arrive independently and in a random order. I can se the model Geom(0.09). E(X) = 1 p = = 11.1 P(X = 4) = (0.91) 3 (0.09) = On average, I expect to have to check jst over 11 s before I find a real message. There s slightly less than a 7% chance that my first real message will be the 4th one I check. Note that the probability calclation isn t new. It s simply Chapter 14 s Mltiplication Rle sed to find P1spam spam spam real2. MATH BOX We want to find the mean (expected vale) of random variable X, sing a geometric model with probability of sccess p. First, write the probabilities: The expected vale is: Let p = 1 - q: Simplify: That s an infinite geometric series, with first term 1 and common ratio q: x Á P1X = x2 p qp q 2 p q 3 p Á E1X2 = 1p + 2qp + 3q 2 p + 4q 3 p + Á = 11 - q2 + 2q11 - q2 + 3q q2 + 4q q2 + Á = 1 - q + 2q - 2q 2 + 3q 2-3q 3 + 4q 3-4q 4 + Á = 1 + q + q 2 + q 3 + Á 1 = 1 - q So, finally... E1X2 = 1 p. Independence One of the important reqirements for Bernolli trials is that the trials be independent. Sometimes that s a reasonable assmption when tossing a coin or rolling a die, for example. Bt that becomes a problem when (often!) we re looking at sitations involving samples chosen withot replacement. We said that whether we find a Tiger Woods card in one box has no effect on the probabilities

4 Independence 391 in other boxes. This is almost tre. Technically, if exactly 20% of the boxes have Tiger Woods cards, then when yo find one, yo ve redced the nmber of remaining Tiger Woods cards. If yo knew there were 2 Tiger Woods cards hiding in the 10 boxes of cereal on the market shelf, then finding one in the first box yo try wold clearly change yor chances of finding Tiger in the next box. With a few million boxes of cereal, thogh, the difference is hardly worth mentioning. If we had an infinite nmber of boxes, there woldn t be a problem. It s selecting from a finite poplation that cases the probabilities to change, making the trials not independent. Obviosly, taking 2 ot of 10 boxes changes the probability. Taking even a few hndred ot of millions, thogh, makes very little difference. Fortnately, we have a rle of thmb for the in-between cases. It trns ot that if we look at less than 10% of the poplation, we can pretend that the trials are independent and still calclate probabilities that are qite accrate. The 10% Condition: Bernolli trials mst be independent. If that assmption is violated, it is still okay to proceed as long as the sample is smaller than 10% of the poplation. STEP-BY-STEP EXAMPLE Working with a Geometric Model People with O-negative blood are called niversal donors becase O-negative blood can be given to anyone else, regardless of the recipient s blood type. Only abot 6% of people have O-negative blood. Qestions: 1. If donors line p at random for a blood drive, how many do yo expect to examine before yo find someone who has O-negative blood? 2. What s the probability that the first O-negative donor fond is one of the first for people in line? Plan State the qestions. Check to see that these are Bernolli trials. Variable Define the random variable. Model Specify the model. I want to estimate how many people I ll need to check to find an O-negative donor, and the probability that 1 of the first 4 people is O-negative. Ç There are two otcomes: sccess = O-negative failre = other blood types Ç The probability of sccess for each person is p = 0.06, becase they lined p randomly. Ç 10% Condition: Trials aren t independent becase the poplation is finite, bt the donors lined p are fewer than 10% of all possible donors. Let X = nmber of donors ntil one is O-negative. I can model X with Geom(0.06).

5 392 CHAPTER 17 Probability Models Mechanics Find the mean. Calclate the probability of sccess on one of the first for trials. That s the probability that X = 1, 2, 3, or 4. E(X ) = L 16.7 P(X 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = (0.06) + (0.94)(0.06) + (0.94) 2 (0.06) + (0.94) 3 (0.06) L Conclsion Interpret yor reslts in context. Blood drives sch as this one expect to examine an average of 16.7 people to find a niversal donor. Abot 22% of the time there will be one within the first 4 people in line. TI TIPS Finding geometric probabilities Yor TI knows the geometric model. Jst as yo saw back in Chapter 6 with the Normal model, commands to calclate probability distribtions are fond in the 2nd DISTR men. Have a look. After many others (Don t drop the corse yet!) yo ll see two Geometric probability fnctions at the bottom of the list. geometpdf(. The pdf stands for probability density fnction. This command allows yo to find the probability of any individal otcome. Yo need only specify p, which defines the Geometric model, and x, which indicates the nmber of trials ntil yo get a sccess. The format is geometpdf(p,x). For example, sppose we want to know the probability that we find or first Tiger Woods pictre in the fifth box of cereal. Since Tiger is in 20% of the boxes, we se p = 0.2 and x = 5, entering the command geometpdf(.2,5). The calclator says there s abot an 8% chance. geometcdf(. This is the cmlative density fnction, meaning that it finds the sm of the probabilities of several possible otcomes. In general, the command geometcdf(p,x) calclates the probability of finding the first sccess on or before the xth trial. Let s find the probability of getting a Tiger Woods pictre by the time we open the forth box of cereal in other words, the probability or first sccess comes on the first box, or the second, or the third, or the forth. Again we specify p = 0.2, and now se x = 4. The command geometcdf(.2,4) calclates all the probabilities and adds them. There s abot a 59% chance that or qest for a Tiger Woods photo will scceed by the time we open the forth box. The Binomial Model We can se the Bernolli trials to answer other qestions. Sppose yo by 5 boxes of cereal. What s the probability yo get exactly 2 pictres of Tiger Woods? Before, we asked how long it wold take ntil or first sccess. Now we want to find the probability of getting 2 sccesses among the 5 trials. We are still talking abot Bernolli trials, bt we re asking a different qestion.

6 The Binomial Model 393 Activity: The Binomial Distribtion. It s more interesting to combine Bernolli trials. Simlate this with the Random Tool to get a sense of how Binomial models behave. NOTATION ALERT: Now pnctation! Throghot mathematics n!, prononced n factorial, is the prodct of all the integers from 1 to n.for example, 4! This time we re interested in the nmber of sccesses in the 5 trials, so we ll call it X = nmber of sccesses. We want to find P1X = 22. This is an example of a Binomial probability. It takes two parameters to define this Binomial model: the nmber of trials, n, and the probability of sccess, p. We denote this model Binom(n, p). Here, n = 5 trials, and p = 0.2, the probability of finding a Tiger Woods card in any trial. Exactly 2 sccesses in 5 trials means 2 sccesses and 3 failres. It seems logical that the probability shold be Too bad! It s not that easy. That calclation wold give yo the probability of finding Tiger in the first 2 boxes and not in the next 3 in that order. Bt yo cold find Tiger in the third and fifth boxes and still have 2 sccesses. The probability of those otcomes in that particlar order is (0.8)(0.8)(0.2)(0.8)(0.2). That s also In fact, the probability will always be the same, no matter what order the sccesses and failres occr in. Anytime we get 2 sccesses in 5 trials, no matter what the order, the probability will be We jst need to take accont of all the possible orders in which the otcomes can occr. Fortnately, these possible orders are disjoint. (For example, if yor two sccesses came on the first two trials, they coldn t come on the last two.) So we can se the Addition Rle and add p the probabilities for all the possible orderings. Since the probabilities are all the same, we only need to know how many orders are possible. For small nmbers, we can jst make a tree diagram and cont the branches. For larger nmbers this isn t practical, so we let the compter or calclator do the work. Each different order in which we can have k sccesses in n trials is called a combination. The total nmber of ways that can happen is written n or k nc k and prononced n choose k. n k = n! k!1n - k2! where n! 1prononced n factorial 2 = n * 1n - 12 * Á * 1 For 2 sccesses in 5 trials, 5 2 = 5! 2!15-22! = 5 * 4 * 3 * 2 * 1 2 * 1 * 3 * 2 * 1 = 5 * 4 2 * 1 = 10. So there are 10 ways to get 2 Tiger pictres in 5 boxes, and the probability of each is Now we can find what we wanted: P1#sccess = 22 = = In general, the probability of exactly k sccesses in n trials is n p k q n-k. k Using this formla, we cold find the expected vale by adding p xp1x = x2 for all vales, bt it wold be a long, hard way to get an answer that yo already know intitively. What s the expected vale? If we have 5 boxes, and Tiger s pictre is in 20% of them, then we wold expect to have = 1 sccess. If we had 100 trials with probability of sccess 0.2, how many sccesses wold yo expect? Can yo think of any reason not to say 20? It seems so simple that most people woldn t even stop to think abot it. Yo jst mltiply the probability of sccess by n. In other words, E1X2 = np. Not flly convinced? We prove it in the next Math Box. The standard deviation is less obvios; yo can t jst rely on yor intition. Fortnately, the formla for the standard deviation also boils down to something simple: SD1X2 = 1npq. (If yo re crios abot where that comes from, it s in the Math Box too!) In 100 boxes of cereal, we expect to find 20 Tiger Woods cards, with a standard deviation of 1100 * 0.8 * 0.2 = 4 pictres. Time to smmarize. A Binomial probability model describes the nmber of sccesses in a specified nmber of trials. It takes two parameters to specify this model: the nmber of trials n and the probability of sccess p.

7 394 CHAPTER 17 Probability Models Binomial probabilities. Do-it-yorseif binomial models! Watch the probabilities change as yo control n and p. BINOMIAL PROBABILITY MODEL FOR BERNOULLI TRIALS: Binom(n, p) n = nmber of trials p = probability of sccess (and q = 1 - p = probability of failre) X = nmber of sccesses in n trials P1X = x2 = n C x p x q n-x, where n C x = Mean: m = np Standard Deviation: s = 1npq n! x!1n - x2! MATH BOX To derive the formlas for the mean and standard deviation of a Binomial model we start with the most basic sitation. Consider a single Bernolli trial with probability of sccess p. Let s find the mean and variance of the nmber of sccesses. Here s the probability model for the nmber of sccesses: x 0 1 P1X = x2 q p Find the expected vale: And now the variance: What happens when there is more than one trial, thogh? A Binomial model simply conts the nmber of sccesses in a series of n independent Bernolli trials. That makes it easy to find the mean and standard deviation of a binomial random variable,y. So, as we thoght, the mean is E1Y2 = np. And since the trials are independent, the variances add: Var1Y2 = Var1X 1 + X 2 + X 3 + Á + X n 2 Var1Y2 = npq E1X2 = 0q + 1p E1X2 = p = pq112 Var1X2 = pq Let Y = X 1 + X 2 + X 3 + Á + X n E1Y2 = E1X 1 + X 2 + X 3 + Á + X n 2 Voilà! The standard deviation is SD1Y2 = 1npq. Var1X2 = 10 - p2 2 q p2 2 p = p 2 q + q 2 p = pq1p + q2 = E1X E1X E1X Á + E1X n 2 = p + p + p + Á + p 1There are n terms.2 = Var1X Var1X Var1X Á + Var1X n 2 = pq + pq + pq + Á + pq 1Again, n terms.2

8 The Binomial Model 395 FOR EXAMPLE Spam and the Binomial model Recap: The commnications monitoring company Postini has reported that 91% of messages are spam. Sppose yor inbox contains 25 messages. Qestions: What are the mean and standard deviation of the nmber of real messages yo shold expect to find in yor inbox? What s the probability that yo ll find only 1 or 2 real messages? I assme that messages arrive independently and at random, with the probability of sccess (a real message) p = = Let X = the nmber of real messages among 25. I can se the model Binom(25, 0.09). E(X) = np = 25(0.09) = 2.25 SD(X ) = 1npq = 125(0.09)(0.91) = 1.43 P(X = 1 or 2) = P(X = 1) + P(X = 2) = a 25 1 b(0.09)1 (0.91) 24 + a 25 2 b(0.09)2 (0.91) 23 = = Among 25 messages, I expect to find an average of 2.25 that aren t spam, with a standard deviation of 1.43 messages. There s jst over a 50% chance that 1 or 2 of my 25 s will be real messages. STEP-BY-STEP EXAMPLE Working with a Binomial Model Sppose 20 donors come to a blood drive. Recall that 6% of people are niversal donors. Qestions: 1. What are the mean and standard deviation of the nmber of niversal donors among them? 2. What is the probability that there are 2 or 3 niversal donors? Plan State the qestion. Check to see that these are Bernolli trials. Variable Define the random variable. Model Specify the model. I want to know the mean and standard deviation of the nmber of niversal donors among 20 people, and the probability that there are 2 or 3 of them. Ç There are two otcomes: sccess = O-negative failre = other blood types Ç p = 0.06, becase people have lined p at random. Ç 10% Condition: Trials are not independent, becase the poplation is finite, bt fewer than 10% of all possible donors are lined p. Let X = nmber of O-negative donors among n = 20 people. I can model X with Binom(20, 0.06).

9 396 CHAPTER 17 Probability Models Mechanics Find the expected vale and standard deviation. E(X ) = np = 20(0.06) = 1.2 SD(X ) = 1npq = 120(0.06)(0.94) L 1.06 P(X = 2 or 3) = P(X = 2) + P(X = 3) = a 20 2 b (0.94) 18 + a 20 3 b(0.06)3 (0.94) 17 L = Conclsion Interpret yor reslts in context. In grops of 20 randomly selected blood donors, I expect to find an average of 1.2 niversal donors, with a standard deviation of Abot 31% of the time, I d find 2 or 3 niversal donors among the 20 people. TI Tips Finding binomial probabilities Remember how the calclator handles Geometric probabilities? Well, the commands for finding Binomial probabilities are essentially the same. Again yo ll find them in the 2nd DISTR men. binompdf( This probability density fnction allows yo to find the probability of an individal otcome. Yo need to define the Binomial model by specifying n and p, and then indicate the desired nmber of sccesses, x. The format is binompdf(n,p,x). For example, recall that Tiger Woods pictre is in 20% of the cereal boxes. Sppose that we want to know the probability of finding Tiger exactly twice among 5 boxes of cereal. We se n = 5, p = 0.2, and x = 2, entering the com- mand binompdf(5,.2,2). There s abot a 20% chance of getting two pictres of Tiger Woods in five boxes of cereal. binomcdf( Need to add several Binomial probabilities? To find the total probability of getting x or fewer sccesses among the n trials se the cmlative Binomial density fnction binomcdf(n,p,x). For example, sppose we have ten boxes of cereal and wonder abot the probability of finding p to 4 pictres of Tiger. That s the probability of 0, 1, 2, 3 or 4 sccesses, so we specify the command binomcdf(10,.2,4). Pretty likely! Of corse p to 4 allows for the possibility that we end p with none. What s the probability we get at least 4 pictres of Tiger in 10 boxes? Well, at least 4 means not 3 or fewer. That s the complement of 0, 1, 2, or 3 sccesses. Have yor TI evalate 1-binomcdf(10,.2,3). There s abot a 12% chance we ll find at least 4 pictres of Tiger in 10 boxes of cereal.

10 The Normal Model to the Resce! 397 The Normal Model to the Resce! Activity: Normal Approximation. Binomial probabilities can be hard to calclate. With the Simlation Tool yo ll see how well the Normal model can approximate the Binomial a mch easier method How close to Normal? How well does a Normal crve fit a binomial model? Check ot the Sccess/ Failre Condition for yorself. Sppose the Tennessee Red Cross anticipates the need for at least 1850 nits of O-negative blood this year. It estimates that it will collect blood from 32,000 donors. How great is the risk that the Tennessee Red Cross will fall short of meeting its need? We ve jst learned how to calclate sch probabilities. We can se the Binomial model with n = 32,000 and p = The probability of getting exactly 1850 nits of O-negative blood from 32,000 donors is a No 1850 b * * calclator on earth can calclate that first term (it has more than 100,000 digits). 1 And that s jst the beginning. The problem said at least 1850, so we have to do it again for 1851, for 1852, and all the way p to 32,000. No thanks. When we re dealing with a large nmber of trials like this, making direct calclations of the probabilities becomes tedios (or otright impossible). Here an old friend the Normal model comes to the resce. The Binomial model has mean np = 1920 and standard deviation 1npq L We cold try approximating its distribtion with a Normal model, sing the same mean and standard deviation. Remarkably enogh, that trns ot to be a very good approximation. (We ll see why in the next chapter.) With that approximation, we can find the probability: P1X = Paz b L P1z L There seems to be abot a 5% chance that this Red Cross chapter will rn short of O-negative blood. Can we always se a Normal model to make estimates of Binomial probabilities? No. Consider the Tiger Woods sitation pictres in 20% of the cereal boxes. If we by five boxes, the actal Binomial probabilities that we get 0, 1, 2, 3, 4, or 5 pictres of Tiger are 33%, 41%, 20%, 5%, 1%, and 0.03%, respectively. The first histogram shows that this probability model is skewed. That makes it clear that we shold not try to estimate these probabilities by sing a Normal model. Now sppose we open 50 boxes of this cereal and cont the nmber of Tiger Woods pictres we find. The second histogram shows this probability model. It is centered at np = = 10 pictres, as expected, and it appears to be fairly symmetric arond that center. Let s have a closer look. The third histogram again shows Binom(50, 0.2), this time magnified somewhat and centered at the expected vale of 10 pictres of Tiger. It looks close to Normal, for sre. With this larger sample size, it appears that a Normal model might be a sefl approximation. A Normal model, then, is a close enogh approximation only for a large enogh nmber of trials. And what we mean by large enogh depends on the probability of sccess. We d need a larger sample if the probability of sccess were very low (or very high). It trns ot that a Normal model works pretty well if we expect to see at least 10 sccesses and 10 failres. That is, we check the Sccess/ Failre Condition. The Sccess/ Failre Condition: A Binomial model is approximately Normal if we expect at least 10 sccesses and 10 failres: np Ú 10 and nq Ú If yor calclator can find Binom(32000,0.06), then it s smart enogh to se an approximation. Read on to see how yo can, too.

11 398 CHAPTER 17 Probability Models MATH BOX It s easy to see where the magic nmber 10 comes from. Yo jst need to remember how Normal models work. The problem is that a Normal model extends infinitely in both directions. Bt a Binomial model mst have between 0 and n sccesses, so if we se a Normal to approximate a Binomial, we have to ct off its tails. That s not very important if the center of the Normal model is so far from 0 and n that the lost tails have only a negligible area. More than three standard deviations shold do it, becase a Normal model has little probability past that. So the mean needs to be at least 3 standard deviations from 0 and at least 3 standard deviations from n. Let s look at the 0 end. We reqire: m - 3s 7 0 Or in other words: m 7 3s For a Binomial, that s: np 7 31npq Sqaring yields: n 2 p 2 7 9npq Now simplify: np 7 9q Since q 1, we can reqire: np 7 9 For simplicity, we sally reqire that np (and nq for the other tail) be at least 10 to se the Normal approximation, the Sccess/Failre Condition. 2 FOR EXAMPLE Spam and the Normal approximation to the Binomial Recap: The commnications monitoring company Postini has reported that 91% of messages are spam. Recently, yo installed a spam filter. Yo observe that over the past week it okayed only 151 of s yo received, classifying the rest as jnk. Shold yo worry that the filtering is too aggressive? Qestion: What s the probability that no more than 151 of s is a real message? I assme that messages arrive randomly and independently, with a probability of sccess (a real message) p = The model Binom(1422, 0.09) applies, bt will be hard to work with. Checking conditions for the Normal approximation, I see that: Ç These messages represent less than 10% of all traffic. Ç I expect np = (1422)(0.09) = real messages and nq = (1422)(0.91) = spam messages, both far greater than 10. It s okay to approximate this binomial probability by sing a Normal model. m = np = 1422(0.09) = s = 1npq = 11422(0.09)(0.91) L P(x 151) = P az b = P(z 2.13) = Among my s, there s over a 98% chance that no more than 151 of them were real messages, so the filter may be working properly Looking at the final step, we see that we need np 7 9 in the worst case, when q (or p) is near 1, making the Binomial model qite skewed. When q and p are near 0.5 say between 0.4 and 0.6 the Binomial model is nearly symmetric and np 7 5 oght to be safe enogh. Althogh we ll always check for 10 expected sccesses and failres, keep in mind that for vales of p near 0.5, we can be somewhat more forgiving.

12 What Can Go Wrong? 399 Continos Random Variables There s a problem with approximating a Binomial model with a Normal model. The Binomial is discrete, giving probabilities for specific conts, bt the Normal models a continos random variable that can take on any vale. For continos random variables, we can no longer list all the possible otcomes and their probabilities, as we cold for discrete random variables. 3 As we saw in the previos chapter, models for continos random variables give probabilities for intervals of vales. So, when we se the Normal model, we no longer calclate the probability that the random variable eqals a particlar vale, bt only that it lies between two vales. We won t calclate the probability of getting exactly 1850 nits of blood, bt we have no problem approximating the probability of getting 1850 or more, which was, after all, what we really wanted. 4 JUST CHECKING As we noted a few chapters ago, the Pew Research Center ( reports that they are actally able to contact only 76% of the randomly selected hoseholds drawn for a telephone srvey. 1. Explain why these phone calls can be considered Bernolli trials. 2. Which of the models of this chapter (Geometric, Binomial, Normal) wold yo se to model the nmber of sccessfl contacts from a list of 1000 sampled hoseholds? Explain. 3. Pew frther reports that even after they contacted a hosehold, only 38% agree to be interviewed, so the probability of getting a completed interview for a randomly selected hosehold is only Which of the models of this chapter wold yo se to model the nmber of hoseholds Pew has to call before they get the first completed interview? WHAT CAN GO WRONG? Be sre yo have Bernolli trials. Be sre to check the reqirements first: two possible otcomes per trial ( sccess and failre ), a constant probability of sccess, and independence. Remember to check the 10% Condition when sampling withot replacement. Don t confse Geometric and Binomial models. Both involve Bernolli trials, bt the isses are different. If yo are repeating trials ntil yor first sccess, that s a Geometric probability. Yo don t know in advance how many trials yo ll need theoretically, it cold take forever. If yo are conting the nmber of sccesses in a specified nmber of trials, that s a Binomial probability. Don t se the Normal approximation with small n. To se a Normal approximation in place of a Binomial model, there mst be at least 10 expected sccesses and 10 expected failres. 3 In fact, some people se an adjstment called the continity correction to help with this problem. It s related to the sggestion we make in the next footnote and is discssed in more advanced textbooks. 4 If we really had been interested in a single vale, we might have approximated it by finding the probability of getting between and nits of blood.

13 400 CHAPTER 17 Probability Models CONNECTIONS This chapter bilds on what we know abot random variables. We now have two more probability models to join the Normal model. There are a nmber of forward connections from this chapter. We ll see the 10% Condition and the Sccess/Failre Condition often. And the facts abot the Binomial distribtion can help explain how proportions behave, as we ll see in the next chapter. WHAT HAVE WE LEARNED? We ve learned that Bernolli trials show p in lots of places. Depending on the random variable of interest, we can se one of three models to estimate probabilities for Bernolli trials: a Geometric model when we re interested in the nmber of Bernolli trials ntil the next sccess; a Binomial model when we re interested in the nmber of sccesses in a certain nmber of Bernolli trials; a Normal model to approximate a Binomial model when we expect at least 10 sccesses and 10 failres. Terms Bernolli trials, if... Geometric probability model Binomial probability model there are two possible otcomes. 2. the probability of sccess is constant. 3. the trials are independent A Geometric model is appropriate for a random variable that conts the nmber of Bernolli trials ntil the first sccess A Binomial model is appropriate for a random variable that conts the nmber of sccesses in a fixed nmber of Bernolli trials. 10% Condition 391. When sampling withot replacement, trials are not independent. It s still okay to proceed as long as the sample is smaller than 10% of the poplation. Sccess/Failre Condition 397. For a Normal model to be a good approximation of a Binomial model, we mst expect at least 10 sccesses and 10 failres. That is, np Ú 10 and nq Ú 10. Skills Know how to tell if a sitation involves Bernolli trials. Be able to choose whether to se a Geometric or a Binomial model for a random variable involving Bernolli trials. Know the appropriate conditions for sing a Geometric, Binomial, or Normal model. Know how to find the expected vale of a Geometric model. Be able to calclate Geometric probabilities. Know how to find the mean and standard deviation of a Binomial model. Be able to calclate Binomial probabilities, perhaps approximating with a Normal model. Be able to interpret means, standard deviations, and probabilities in the Bernolli trial context.