www.downloadslide.com 3.4 The Binomial Probability Distribution 105 The binomial probability distribution has many applications because the binomial experiment occurs in sampling for defectives in industrial quality control, in the sampling of consumer preference or voting populations, and in many other physical situations. We will illustrate with a few examples. Other practical examples will appear in the exercises at the end of this section and at the end of the chapter. EXAMPLE 3.7 Solution Suppose that a lot of 5000 electrical fuses contains 5% defectives. If a sample of 5 fuses is tested, find the probability of observing at least one defective. It is reasonable to assume that Y , the number of defectives observed, has an approximate binomial distribution because the lot is large. Removing a few fuses does not change the composition of those remaining enough to cause us concern. Thus, ( 5 P(at least one defective) = 1 − p(0) = 1 − p 0) 0 q 5 = 1 − (.95) 5 = 1 − .774 = .226. Notice that there is a fairly large chance of seeing at least one defective, even though the sample is quite small. EXAMPLE 3.8 Solution Experience has shown that 30% of all persons afflicted by a certain illness recover. A drug company has developed a new medication. Ten people with the illness were selected at random and received the medication; nine recovered shortly thereafter. Suppose that the medication was absolutely worthless. What is the probability that at least nine of ten receiving the medication will recover? Let Y denote the number of people who recover. If the medication is worthless, the probability that a single ill person will recover is p = .3. Then the number of trials is n = 10 and the probability of exactly nine recoveries is ( ) 10 P(Y = 9) = p(9) = (.3) 9 (.7) = .000138. 9 Similarly, the probability of exactly ten recoveries is ( ) 10 P(Y = 10) = p(10) = (.3) 10 (.7) 0 = .000006, 10 and P(Y ≥ 9) = p(9) + p(10) = .000138 + .000006 = .000144. If the medication is ineffective, the probability of observing at least nine recoveries is extremely small. If we administered the medication to ten individuals and observed at least nine recoveries, then either (1) the medication is worthless and we have observed a rare event or (2) the medication is indeed useful in curing the illness. We adhere to conclusion 2.
www.downloadslide.com 106 Chapter 3 Discrete Random Variables and Their Probability Distributions A tabulation of binomial probabilities in the form ∑ a y=0 p(y), presented in Table 1, Appendix 3, will greatly reduce the computations for some of the exercises. The references at the end of the chapter list several more extensive tabulations of binomial probabilities. Due to practical space limitations, printed tables typically apply for only selected values of n and p. Binomial probabilities can also be found using various computer software packages. If Y has a binomial distribution based on n trials with success probability p, P(Y = y 0 ) = p(y 0 ) can be found by using the R (or S- Plus) command dbinom(y 0 ,n,p), whereas P(Y ≤ y 0 ) is found by using the R (or S-Plus) command pbinom(y 0 ,n,p). A distinct advantage of using software to compute binomial probabilities is that (practically) any values for n and p can be used. We illustrate the use of Table 1 (and, simultaneously, the use of the output of the R command pbinom(y 0 ,n,p)) in the following example. EXAMPLE 3.9 The large lot of electrical fuses of Example 3.7 is supposed to contain only 5% defectives. If n = 20 fuses are randomly sampled from this lot, find the probability that at least four defectives will be observed. Solution Letting Y denote the number of defectives in the sample, we assume the binomial model for Y , with p = .05. Thus, P(Y ≥ 4) = 1 − P(Y ≤ 3), and using Table 1, Appendix 3 [or the R command pbinom(3,20,.05)], we obtain P(Y ≤ 3) = 3∑ p(y) = .984. y=0 The value .984 is found in the table labeled n = 20 in Table 1, Appendix 3. Specifically, it appears in the column labeled p = .05 and in the row labeled a = 3. It follows that P(Y ≥ 4) = 1 − .984 = .016. This probability is quite small. If we did indeed observe more than three defectives out of 20 fuses, we might suspect that the reported 5% defect rate is erroneous. The mean and variance associated with a binomial random variable are derived in the following theorem. As you will see in the proof of the theorem, it is necessary to evaluate the sum of some arithmetic series. In the course of the proof, we illustrate some of the techniques that are available for summing such series. In particular, we use the fact that ∑ y p(y) = 1 for any discrete random variable.