All statistics are based on probability. Probability is just a formal way of referring to chance. If, for example, we were to flip a coin, we would know that there is some chance that it will come up heads and, by implication, some chance that it will come up tails. We know, in general, that this chance is fifty-fifty for either heads or tails occurring. But we also know that there is no certainty at all to the way the coin will come up. If the coin is flipped in what any of us would think of as the normally appropriate way, we are pretty sure that the resulting head or tail will have come up totally by chance or, in statistical terms, randomly. This random occurrence of heads or tails makes the result of a coin flip conform to what statisticians call a stochastic process. Simply put, a stochastic process is one for which any result has a certain probability of occurrence. In this case, the probability of any result of the flip of a single coin (either heads or tails) is 0.5.

Any discussion of probability requires agreement on some initial concepts and definitions. The following sections provide working definitions for probability concepts.

An observation to which a probability can be assigned is often referred to as an event. If we flip a coin, we might reasonably think of that as an event (that is, the flip). But in this chapter we will actually think of the event as the viewing of the results of the flip. To understand how we will consider this, think of the story of the three umpires—the traditionalist, the pragmatist, and the existentialist. The traditionalist umpire, in calling balls and strikes, says, “I call ’em as they are.” The pragmatist says, “I call ’em as I see ’em.” The existentialist says, “They ain’t nothin’ until I call ’em.” We will take the existentialist view. The result of a coin flip is nothing until we view it and “call” the result.

Independent Events If we flip a coin twice, it is generally assumed that the outcome of the first flip will have no bearing on the outcome of the second flip. The first event does not influence the second event in any way. In statistical terms, it is said that the two events (flip one and flip two) are independent of each other. When we flip the coin the first time, the probability of our getting either a head or a tail is 0.5. When we flip the coin the second time, the probability of either a head or a tail is again 0.5. There are four possible outcomes of the flip of two coins: HH, HT, TH, or TT. The probability of any one of these four outcomes is 0.25, because each is equally likely and one divided by four is 0.25. But we can also arrive at the probability of any one of these four outcomes by the rule that says the probability of any outcome of two independent events is the multiplication of the probability of each separate event. Thus, the probability of the outcome HH is 0.5 times 0.5, which is 0.25, which is also the probability of the outcome TH, HT, or TT. This rule is referred to as the simple multiplication rule.

Sample Space The two possible outcomes from the flip of a coin—heads or tails—are frequently called the sample space for this process. The sample space refers to all the possible outcomes of a particular process. The sample space for two flips of a coin includes the outcomes HH, HT, TH, and TT. For five flips of a coin, the sample space includes 2⁵, or thirty-two, separate outcomes. As we have seen, each of these individual outcomes has the same probability (0.03125) of occurring. But it is often less the specific outcome than the mix of heads and tails that will interest us in the flip of five coins.

FIGURE 5.1 Possible combinations of five coin flips

Figure 5.1 shows the thirty-two ways that the five flips of a coin can come out in terms of the number of heads realized. As the table shows, there is one way to obtain five heads, five ways to obtain four heads, and so on. In five flips of a coin, each of the thirty-two outcomes is mutually exclusive of all others. Being mutually exclusive means, for example, that the result HHHHH and the result HTTHT cannot both occur. One can occur or the other can occur, but the occurrence of one precludes the occurrence of the other (and, similarly, all other outcomes).

FIGURE 5.2 Venn diagram for two mutually exclusive events

There are two types of probability to consider. The first of these is the probability associated with the flip of a coin or, for example, with the roll of a die or the draw of a card from a deck. The nature of the object or objects upon which the random process is based determines the probability of the outcome. The probability of the outcome of a head after the flip of a coin is 0.5 given the nature of the coin (it has only two sides). The probability that a two will come up on the roll of a fair six-sided die is approximately 0.1667 (1/6) given the nature of the die. The probability of drawing an ace from the top of a well-shuffled deck is approximately 0.0769 (4/52) given the nature of the deck of cards (there are four aces in a fifty-two-card deck). This type of probability is called a priori probability. Virtually all games of chance are based on a priori probabilities.

Frequency of Occurrence and Empirical Probability What we can rely on to gauge this probability of occurrence is some historical record of persons arriving at the emergency room. Suppose we have a record of all the arrivals at the emergency room for the past six months, during which time there were 7,320 total emergency room visits (about forty per day). We also have information about whether the visit was for a true emergency as assessed by the attending physician, physician’s assistant, or nurse. Suppose the record indicates that 4,729 of the visits were true emergencies. Now we can say that for the past six months, the probability that a person coming to the emergency room came for a true emergency was 0.646 (4,729/7,320). Based on this, and assuming that the future is like the past, we could predict that when the next person arrives at the clinic, that person has about a 0.65 probability of having a true emergency and about a 0.35 probability of having a nonemergency problem. This type of probability is known as empirical probability.

FIGURE 5.3 Children ever born to 2,556 women

Sequential Events and Empirical Probabilities Empirical probabilities work just like a priori probabilities in regard to sequential events. For example, if we looked at the record of the sequential arrival of any two people at the emergency room, we would see four possible events. If we classify true emergencies as E and nonemergency conditions as O (other), then the results of the arrival of two sequential people could be EE, EO, OE, and OO. Now, however, despite the fact that there are four possible outcomes, the probability of each outcome is not 0.25. If we do assume for the moment that the outcome of the arrival of the first person (that is, E or O) does not influence the outcome of the arrival of the second person, then we can say that the arrival outcomes are independent of one another. Under the assumption of independence, the probability that the outcome will be EE is 0.65 × 0.65, or 0.42. The probability of EO is 0.65 × 0.35, or 0.23. The probability of the outcome OE is the same as that of EO, and the probability of the outcome OO is 0.35 × 0.35, or 0.12. Because these four mutually exclusive outcomes represent the entire sample space, the reader can confirm that the probabilities for the four sum to 1. Figure 5.4 displays the probability tree associated with emergency and nonemergency situations. Per the probability tree, marginal probabilities are listed first, followed by conditional probabilities and finally joint probabilities.

FIGURE 5.4 Sequential events

Independence and Empirical Probabilities With empirical probabilities, though, unlike the case with a priori probabilities, we may not always be justified in viewing many events as independent. It is likely that several persons arriving at an emergency room in a sequential manner may be dependent on one another in regard to their emergency or nonemergency status. If, for example, there occurred a large highway accident that sent a number of persons to emergency rooms, several persons arriving at the same time might all be related (and hence not independent) emergencies. In fact, one of the most important capabilities of statistics is the ability to assess whether events are independent of or dependent on one another. The concept of independence is one that we will return to over and over in the course of this book.

Empirical Probability and Probability Distributions In many instances, data of interest to health workers can be viewed only from the standpoint of empirical probability. Whether an arrival at an emergency room is an emergency or not can be understood only as an empirical probability. But, interestingly, once we know what the overall probability of being an emergency is, and if we assume that the arrival of an emergency is independent of whether the last person arriving was an emergency, we can apply a priori probability assumptions to the arrival of several people at a time. In other words, those arrivals could be modeled using a probability distribution, and the a priori assumption we apply is that those arrivals are from a binomial distribution. The binomial distribution is discussed in detail later in this chapter. In fact, numerous distributions can be used to model data depending on the specific character or shape of the data. Another very important distribution is the normal distribution. The normal distribution is important in that one of the most important assumptions for statistics is that both discrete numerical data and continuous numerical data will somehow follow, a priori, a normal distribution. Normal distributions are discussed further in this chapter, and also in chapter six.

To continue with the discussion of probability, it will be useful to distinguish three different types of probabilities: marginal probabilities, joint probabilities, and conditional probabilities. Marginal probabilities are those associated with a single event. “Joint probability” refers to the simultaneous occurrence of two or more types of events. “Joint probabilities” and “conditional probabilities” refer to the outcomes of two different types of events that may or may not be independent of each other.

The probability (proportion) shown in Figure 5.3 is often referred to as marginal probability . Marginal probability is the probability of the occurrence of a single outcome. The probability that any single woman selected at random from among the 2,556 women will have no children is the number of women with no children divided by the total number of women. “Marginal probability” refers to the probability of an occurrence of any single outcome. By the same token, whether the next person appearing at the emergency room arrives with a true emergency or a nonemergency condition is also a single outcome (or event if we are counting only that one person). The marginal probability for true emergencies, according to the discussion earlier in this chapter, is 0.65, and for nonemergency conditions it is 0.35.

“Marginal probability” refers to the outcome of a single event or type of event (the birth of children, the arrival at an emergency room). “Joint probability” refers to the simultaneous occurrence of two or more types of events. Return to the example of persons arriving at an emergency room. Figure 5.5 shows the first twenty observations in a file that contains records of 7,320 emergency room visits. Column C shows whether the visit was for an emergency or not. Column B is labeled Shift. The first shift (labeled 1st) is from 7 A.M. to 3 P.M. The second shift (labeled 2nd) is from 3 P.M. to midnight, and the third shift (labeled 3rd) is from midnight to 7 A.M.

FIGURE 5.5 First twenty observations in an emergency room visit file

Building a Joint Probability Table Using a Spreadsheet In this example, we might, in addition to determining whether the visit was for a true emergency or a nonemergency condition, determine when during the day the arrival took place. The pivot table option in Excel was used to produce Figure 5.6, which shows the joint occurrence of a true emergency or a nonemergency condition and the shift during which the visit took place. It should be easy to recognize that Figure 5.6 is an example of a contingency table. A contingency table is one that shows the simultaneous occurrence of two or more events. In this case, the events are whether the arrival at the emergency room is or is not an emergency and the shift during which the arrival occurred.

FIGURE 5.6 Contingency table of shift and emergency status

FIGURE 5.7 Joint probabilities for shift and emergency status

Mutual Exclusivity and the Simple Addition Rule It can be confirmed from the table that the marginal probability (p) that the visit was for a true emergency is equal to the sum of the joint probabilities for the first, second, and third shifts. This can be seen in Equation 5.1. Equation 5.1 is referred to as the simple addition rule. The simple addition rule can be applied here since the first, second, and third shifts are mutually exclusive (i.e., they cannot occur at the same time).

Mutual Exclusivity and the Addition Rule Thus far, the discussion has concerned a joint probability that is represented by the conjunction “and.” It is also possible to discuss a joint probability that is represented by the conjunction “or.” The joint probability “or” would be stated in regard to true emergencies and time of day. For example, a visit was for a true emergency or it occurred during the first shift. The joint probability “or” is the sum of the appropriate marginal probabilities minus the appropriate joint probability. At times this is referred to as “avoiding the double count.” We must subtract the joint probability of the two events if those two events are not mutually exclusive. In our example, a visit for a true emergency is not mutually exclusive of time of day. Figure 5.8 is a Venn diagram representing the events from our example, true emergency (TE) and first shift (FS). Notice that the two circles overlap. The overlap represents the intersection of the two events or the joint probability. The overlap indicates that these two events are not mutually exclusive.

FIGURE 5.8 Venn diagram of two events that are not mutually exclusive

FIGURE 5.9 Joint probability “or” for shift and emergency status

Calculating Joint Probability “or” Values in a Spreadsheet Figure 5.9 shows all the joint probability “or” values for arriving during one of three shifts or for coming for an emergency or for a condition that is not an emergency. The formula line shows the calculation of the joint probability p(True or First) as B$13+$E11-B11. This formulation lets us drag the formula in cell B17 to the other five cells and generates the joint probability “or” for each of the cells in the table. The joint probabilities “or” for each row and column have been summed in cells B19:D19 and in cells E17 and E18. At first glance, these numbers mean nothing, because all probability values must be equal to or less than 1 to have meaning. However, the sum of the row probabilities or the sum of the column probabilities can be summed to the total of the joint probabilities “or” in cell E19. For a two-by-three table, such as the one shown in Figure 5.9, the sum of the joint probabilities “or” will always be 4. Four turns out to be the number of rows in the figure plus the number of columns minus 1. In fact, for any contingency table of any size, the sum of all joint probabilities “or” is always equal to the number of rows plus the number of columns minus 1.

The discussion of probabilities given thus far has involved the development of probabilities with essentially no prior knowledge of events. The discussion has focused, for example, on the probability that a visit to the emergency room would be for a true emergency and would occur during the first shift. Now the discussion turns to another point of probability—the probability of one event’s occurring if it is known that another event has occurred. For example, suppose we know that an emergency room visit has taken place during the third shift. The task then might be to determine the probability that the visit would be a true emergency, given that it is already known that it took place during the third shift. We might also ask, What is the probability that the visit was an emergency conditional on its having taken place during the third shift? This is known as conditional probability.

Bayes’ Theorem and Conditional Probability The calculation of conditional probabilities can be taken directly from our discussion of marginal and joint probabilities and may be shown as given in Equation 5.3. In Equation 5.3, p(A|B) is read as “the probability of A given B,” which means the probability of A is conditional on B’s having occurred, or having been known. This is also known as Bayes’ Theorem.

FIGURE 5.10 Conditional probabilities for arrival during any shift

Conditional Probabilities and Data Frequencies Conditional probabilities can also be calculated directly from data frequencies. It should be noted that the conditional probability of the visit to the emergency room being for a true emergency, given that it took place during the first shift, is just the number of visits for true emergencies taking place during the first shift, divided by all the visits taking place during the first shift. If we look at Figure 5.6, for example, we can see that there were 2,652 visits during the first shift, of which 1,504 were for true emergencies. If we divide 1,504 by 2,652, the result is 0.567, which is the same as what we found using the formula in Equation 5.4, or as calculated in Figure 5.10.

Conditional Probability and Independent Events This introduction of the concept of conditional probability provides another opportunity to talk about the notion of independent events and independence. Earlier, independence was defined as being a situation in which the occurrence of one event had no relation to the occurrence of a second event. In particular, if a coin is flipped twice, the second flip is generally assumed to be independent of the first because the result of the first will not influence the result of the second. If the first is a head, we should not expect that a second flip will be any more likely a tail than a head.

Conditional Probability Tables via Spreadsheet to Test Independence If we had been dealing with a table of two rows and two columns, any conditional probability that did not equal the marginal probability would have forced all conditional probabilities to be different from the marginal probabilities. The nature of a two-by-two table is such that if p(A|B) ≠ p(B) for any individual conditional probability in the table, the principle of independence will hold for all other conditional probabilities in the table. In a table with more than four internal cells, we cannot make that claim.

FIGURE 5.11 Conditional probabilities for high- and low-income women and number of children

Mathematical Versus Statistical Independence It is important to point out, however, that the concept of independence discussed thus far is mathematical independence. If Equation 5.5 does not hold, then two events are not mathematically independent. By this definition, the two events, arrival for an emergency and arrival during a particular shift, are not independent. But the conditional probabilities for arrival for an emergency or for a nonemergency condition, given a shift during the day, are not much different from the marginal probabilities. As Figure 5.10 shows, the conditional probability of arriving for an emergency during any particular shift is in every case no more than 0.1 different from the marginal probability. Mathematically speaking they are different; however, viewed another way, are they statistically different? Could we think of those probabilities in terms of statistical independence? The answer to this question is yes. Chapter eight discusses statistical tests for categorical data. These test allow us to determine whether the differences between conditional and marginal probabilities can be considered statistically independent.

As has been discussed thus far, a person comes to the emergency room for one of two reasons: either for an emergency or for a nonemergency reason. Because arrival can be for one of two reasons, this can be considered a binary event. The occurrence of binary events generally follows a known distribution, called the binomial distribution. Consider, again, a most common binary event—the flipping of a coin. The result can be either a head or a tail, and either outcome is equally likely (i.e., either outcome has a 0.5 probability of occurrence). But suppose we are flipping the coin five times. Figure 5.1 showed the number of different heads that could be obtained in five flips of a coin. That table indicates that the probability of getting all heads was 0.03125, the probability of getting four heads was 0.15625, the probability of getting three heads was 0.3125, and the probability of getting two one or no heads were exactly the same probabilities in reverse. These probabilities for five flips of a coin follow the binomial distribution.

Applying Binary Logic to Flips of a Coin But is it possible to generalize the information found in Figure 5.12 to any number of coin flips, or, for that matter, to any binary events such as the arrival for an emergency or nonemergency condition? The answer is yes. To do so, let us first consider the probability of getting any number of heads in five flips of the coin. Let us say that we flip a coin five times and we get the outcome HHTHT. This is outcome number six in Figure 5.12. It can then be asked, “What is the probability of getting that outcome?”

FIGURE 5.12 All possible outcomes of the flip of a coin five times

where p(I) is the probability of the event I (for example the probability of a head in one flip of a coin) andis the probability of any single outcome of x events in n tries (for example the outcome HHTHT in five flips of a coin). For example, the probability of the specific pattern of heads and tails HHTHT is

Applying Binary Logic to Emergency and Nonemergency Visits For the flip of a coin where the probability of heads and one minus the probability of heads are the same (0.5), the result of Equation 5.6 is exactly the same for any single value of x (that is, for any single number of heads, such as 1 head, 2 heads, and so on). If we consider again the visit of people to an emergency room and use the empirical probabilities shown in Figure 5.6, p(I) and 1—p(I) are not equal (e.g., p(E) = 0.65 and p(O) = 1- 0.65 = 0.35). Consequently, the result of the application of Equation 5.6 will not be the same for every different number of true emergencies (x) and nonemergency conditions. Figure 5.13 shows all the possible outcomes of the visit to an emergency room of five people and their emergency or nonemergency status. We assume that each visit is independent of all other visits. The probability of five emergencies is given as 0.113. You can confirm that this is 0.646⁵ or p(I)⁵ × (1 - p(I))^(5-5). The equation in cell H2 would resemble this: =(0.646^(G2))*((1-0.646)^(5-G2)). The probability of four emergencies out of five (for example outcome number 2 or outcome number 3) is 0.062. It is possible to confirm that this is 0.646⁴ × 0.354¹ or p(I)⁴ × (1 - p(I))^(5-4), as expressed in Equation 5.6.

But there is a less tedious way of getting the result in Figure 5.14 than by enumerating every possible combination. As both Figures 5.13 and 5.14 show, the probability of getting any single number of emergencies is exactly the same for each one of the ways in which that number of emergencies can be obtained. For example, the probability of seeing three emergencies out of five arrivals at the emergency room is 0.034. This is true for any of the ten ways to see three emergencies. Consequently, if there is a way to determine the number of ways that one can find three emergencies out of five visits, it would be possible simply to multiply this number times the probability of finding three emergencies in any single way (Equation 5.6).

Combinatorial Formulas It turns out that there is a formula for finding the number of ways that one can get three emergencies out of five visits. More generally we wish to find x emergencies (I) out of n total observations. That formula is the general formula for the number of combinations out of any number of tries, as is shown in Equation 5.7. Equation 5.7 says that the total number of combinations, (C), for finding n emergencies (I or the Index event) out of n observations is n factorial divided by the quantity x factorial times (n - x) factorial.

FIGURE 5.13 All possible outcomes of five emergency room visits

FIGURE 5.14 Probabilities of number of visits that are actual emergencies

where C is the number of different ways to get x results of I out of n observations. (Three emergencies out of five persons coming to the emergency room, for example.)

Factorials in Excel: The =FACT() Function The term “factorial” means to multiply the number to which the factorial refers by every number less than it in the number sequence. So, for example, 5! is 5 × 4 × 3 × 2 × 1, or 120. The Excel =FACT() function will provide the value of factorials up to 170! (about 7.257E+306 or 7,257 followed by 303 zeros). Now if we wish to know the probability of seeing, for example, x emergencies in the next n persons who come to the clinic, the complete formula is as given in Equation 5.8.

whereis the total probability of n results of I out of x observations (the binomial probability).

FIGURE 5.15 Probabilities of number of visits using formulas

FIGURE 5.16 Formulas used for calculations of probabilities

Binomial Probabilities in Excel: The =BINOMDIST() Function Figure 5.15 was calculated using the formulas in Equations 5.6, 5.7, and 5.8. Excel actually provides a function that produces an easier way to determine binomial probabilities. This is the =BINOMDIST() function. The =BINOMDIST() function takes four arguments. These are the number of emergency visits (5, 4, 3, 2, 1, or 0), the number of visits observed (five), the probability of an emergency (0.646), and a 0 or 1 to indicate whether the value to be determined is the actual probability or the cumulative probability. Figure 5.17 shows the result of using the =BINOMDIST() function to calculate binomial probabilities. The values in column B have been calculated by the formula shown in the formula bar. The values in column C have been calculated by dragging the equation from column B to column C and changing the 0 in the final argument to a 1.

FIGURE 5.17 The =BINOMDIST() function

Binomial Applications to the Health Care Professional To this point there has been ample time spent on the development of the binomial distribution. Therefore it is appropriate to relate the binomial distribution to the health care professional. The binomial distribution can give the probability for the occurrence of any event that can have two outcomes. Let’s continue with emergency room visits example. Suppose a hospital emergency room administrator has data over the past year showing that exactly 20,278 true emergencies were seen in the emergency room out of 31,390 people who sought care from the emergency room. Suppose further that the emergency room administrator is pretty certain that most visits to the emergency room are independent events (i.e., they are not related to one another such as patients from a train wreck or a multiple-car pileup). Using the binomial event logic, the emergency room administrator can figure out how many emergencies she might expect during an eight-hour shift.

FIGURE 5.18 Binomial distribution for emergencies in an eight-hour shift

Answering the Question of Correctly Documented Medicare Claims In chapter one there is a brief discussion of the problem of incorrectly documented Medicare claims at the Pentad Home Health Agency. The agency drew a random sample of eighty records from their files, and during an audit of the records it was determined that only 75 percent of them had been correctly documented. The agency believed that any fewer than 85 percent correctly documented would lead to real reimbursement difficulties with the Medicare administration. They were concerned that they should initiate a training activity for staff to ensure that at least 85 percent of claims would be correctly documented in the future. But they were not completely convinced that just because only 75 percent were correctly documented in the audit, fewer than 85 percent of all records had been incorrectly documented. The training activity was not going to be cheap, and it would be a real waste to undertake the training if it was not really needed. What information could they call on to make a decision as to whether to undertake training?

Calculating the Binomial Probability of Correctly Documented Claims The first way would be to ask, If we have found 75 percent of our sample correctly documented, what is the probability that 85 percent of the population from which the sample came was correctly documented? Considering eighty records, 85 percent would be sixty-eight records. So we can use the binomial distribution equations given in Equations 5.6, 5.7, and 5.8, or the =BINOMIAL(67,80,0.75,1) function, to determine that the cumulative binomial probability of obtaining a value of sixty-seven or fewer correctly documented in any random sample of eighty records is 0.978. We can also determine that 1 - (=BINOMIAL(67,80,0.75,1)) is 0.022. Therefore there is about a 2.2 percent chance that the true proportion of correctly documented records is 85 percent or more when the sample proportion is 75 percent.

Visualizing the Binomial Distributions of Correctly Documented Claims Another way of considering either of these alternatives might be by visualizing the actual binomial distributions for each. Figure 5.19 shows the binomial distribution for 75 percent correct out of eighty observations (the light gray distribution on the left) and the binomial distribution for 85 percent correct out of eighty observations (the black distribution on the right). These were generated using the =BINOMDIST(x,80,p,0) function, where x was all values from fifty to eighty and p was either 0.75 or 0.85. If we look first at the light gray distribution (p = 0.75), we can see that only a small proportion of that distribution (actually 2.2 percent) is to the right of the vertical black line that marks the number sixty-eight (85 percent of eighty). But if we look at the black distribution (p = 0.85), we can see that only a small proportion of that distribution (actually, 1.3 percent) is to the left of the vertical black line that marks the number sixty (75 percent of eighty). This is graphic evidence that it is very unlikely that the true proportion of correctly documented records is 85 percent when the sample finds 75 percent.

FIGURE 5.19 Binomial distributions for 0.75 and 0.85 correct

The Poisson distribution, like the binomial distribution, is a discrete distribution. It takes on values for whole numbers only. But whereas the binomial distribution is concerned with the probability of two outcomes over a number of trials, the Poisson distribution is concerned with the number of observations that will occur in a small amount of time or over a region of space. We have been discussing the probability of seeing a certain number of emergencies out of a number of persons who arrive at an emergency room. The binomial distribution describes these probabilities. But if we consider the actual arrival at the emergency room of anyone, emergency or not, the Poisson distribution is more likely to describe these probabilities.

Arrivals at the Emergency Room: Application of the Poisson Distribution Consider again the emergency room that deals with twenty-nine people, on average, during an administrator’s eight-hour shift. The administrator can figure out that this is an average of just about 0.9 persons—or just less than one person—every fifteen minutes. She also knows that it takes about fifteen minutes to go through all the administrative paperwork—checking insurance coverage, inquiring about previous visits, and so on—entailed in getting a person into the system. She knows that she will have to be prepared to deal with about one person every fifteen minutes, but she also knows that on occasion several people arrive within the same fifteen-minute time interval, even when they arrive for different reasons. She would like to know, then, how often she will have to be prepared to deal with two people, or three or four, in any fifteen-minute interval. She can determine this with the Poisson distribution.

The =POISSON() Function in Excel Figure 5.20 shows the Poisson-predicted probabilities for the number of arrivals in any fifteen-minute period. The probabilities are calculated using the =POISSON() function, which takes three arguments. These are the number of arrivals (A2 in the formula line), the average number of arrivals during the period of time under consideration ($B$11 in the formula line, and generally termed l in the context of the Poisson distribution), and a zero to indicate that the actual probability is desired.

FIGURE 5.20 Poisson distribution of emergency room arrivals in fifteen-minuteintervals

In the section on the binomial distribution, we spent a good deal of time showing where the binomial distribution comes from. This was for two reasons:

FIGURE 5.21 Calculated Poisson distribution of emergency room arrivals in fifteen-minute intervals

Figure 5.21 shows the calculated values of the Poisson distribution for emergency room arrivals in fifteen-minute intervals. Column B gives the value of δ^x, where x represents zero to six arrivals. Column C is simply the value of e (2.718) raised to the power of -δ. Column D is the factorial value for each number of arrivals, zero to six. Column E uses the formula in Equation 5.9 to replicate the values in column B in Figure 5.20.

FIGURE 5.22 Calculated Poisson distribution of emergency room arrivals: Excel formulas

FIGURE 5.23 Poisson distribution for gloves that are not usable in a box of one hundred

The last topic that will be taken up in this chapter is the normal distribution. Both the binomial distribution and the Poisson distribution describe discrete numerical variables. That is, they give the probability of a certain number of true emergencies among visitors to an emergency room. Discrete events are individually distinct and take on a finite or countable number of values.

The normal distribution, however, provides probabilities for continuous numerical variables. Height of adult males, for example, is a continuous variable. Body temperature is another continuous variable, as is pulse rate or blood pressure. The amount of time spent in a physician’s waiting room or with the physician is a continuous variable. All of these variables can be determined not by counting, which can be done with a discrete variable, but by measurement. In general, it is reasonable to say that a continuous numerical variable is the result of some measurement—say, with a watch, a ruler, or a scale.

FIGURE 5.24 A normal distribution

To briefly introduce the normal distribution, consider the chart shown in Figure 5.24. This chart represents the normal distribution. The normal distribution is discussed in detail in chapter six, but several points about it are noteworthy here. First, the normal distribution has often been called a bell-shaped curve because it looks like the cross-section of a bell. Second, because the normal distribution looks like a bell, both sides of the distribution have exactly equal probabilities. In other words, the normal distribution is symmetrical. For example, if the probability of being at the point -1 on the horizontal scale is 0.024, then the probability of being at 1 on the horizontal scale is also 0.024. Moreover, the probability of being between -3 and 0 is exactly the same as being between 0 and 3.

In Figure 5.24, the horizontal axis is given as standard deviations. The meaning of the term “standard deviation” is discussed in detail in chapter six. So, for now we will use the term in the context of the normal distribution. It is sufficient to say that in any normal distribution, approximately 68 percent of all observations will be within ±1 standard deviation of the center of the distribution. Let’s assume we have a normally distributed data set of average height of adult males. Consequently, then it can be expected that in a large sample of adult males, 68 percent will have heights no less than one standard deviation below the average and no more than one standard deviation above the average. Furthermore, if the average height of adult males is normally distributed, 95 percent of all adult males will have heights between ±2 standard deviations of the average. Following this logic, 99 percent will have heights between ±3 standard deviations of the average.

The average height of adult males in the United States is about sixty-nine inches. Let us say that virtually all adult U.S. males are at least sixty inches tall and no more than seventy-eight inches tall. Of course, there are some U.S. adult males—professional basketball players, for example—who are notably taller than seventy-eight inches. In fact, this is also a characteristic of normal distributions. Normal distributions have no upper or lower limits. If the range of adult human height is sixty to seventy-eight inches and if height is normally distributed, then we could say that the standard deviation of height is about three inches (78 - 60 = 18, 18/6 = 3). Consequently, we could say that 68 percent of all U.S. males are between sixty-six and seventy-two inches tall (e.g., 69 ± 1*3 = 66 and 72). Furthermore, we could say that 95 percent of all adult males are between sixty-three and seventy-five inches tall (e.g., 69 ± 2*3 = 63 and 75). Finally, 99 percent are between sixty and seventy-eight inches tall (e.g., 69 ± 3*3 = 60 and 78).