Idea Transcript
CHAPTER
13
Binomial Distributions 13.1 Binomial Distributions 13.2 Common Errors 13.3 Selected Exercise Solutions
13.1 Binomial Distributions We begin by demonstrating how to compute various probabilities for a given binomial distribution, the most common type of distribution for counts and proportions.. To do so, we will need the binompdf( and binomcdf( commands from the DISTR menu. Actual location of these depends on the calculator model. 73
74 Chapter 13 – Binomial Distributions
Binomial Probabilities For a binomial distribution X ~ B(n, p) , we compute the probability of exactly k successes, P(X = k ) , by entering the command binompdf(n,p,k). The probability P( X ≤ k ) = P (0 ≤ X ≤ k ) of at most k successes is computed with the command binomcdf(n,p,k). The probability of there being at least k successes is given by P ( X ≥ k ) = 1 − P( X ≤ k −1) , and is computed with the command 1-binomcdf(n,p, k-1). The following examples demonstrate these calculations. TI-89 calculators explicitly ask for the low and high ends of interest to be included in the calculation, so there is no need for any subtraction. Example 13.4 Inheriting blood type. The blood types of successive children born to the same parents are independent, and have fixed probabilities that depend on the genetic makeup of the parents. Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have five children, what is the probability that exactly 2 of them have type O blood? Let X be the number of children with type O blood. X~B(5, 0.25). We need to find P(X = 2). TI-83/84 Solution. Since we are interested in exactly two of these children, we use command binompdf(5,.25,2) to obtain P(X = 2) ≈ 0.2637, a little more than 26%.
TI-89 Solution. In the Stats/List Editor application, press ‡ for the Distr menu. Arrow down to find Binomial Pdf. Press Í to select it. You will have a dialog box like the one below that explicitly asks for n, p, and k. Press Í to perform the calculation.
Example 13.5 Inspecting CDs. The results X of CDs with defective copy protection in Example 13.3 has approximately the binomial distribution with n = 10 and p = 0.1. What is the probability of at most one defective CD in the sample of 10? Solution. For the probability of at most 1 defective, we could use binompdf (Binomial Pdf) twice to find the probabilities of exactly 0 and 1 defectives, and then
Normal Approximation for a Binomial 75
add the results, but it is more efficient to use binomcdf (Binomial Cdf) and let the calculator do the addition for us. The chance of at most one defective in our sample of 10 is about 73.61%.
Example Free Throws. Suppose a basketball player makes 75% of his free throws. In one particular game, he missed 5 of 12 attempts. Is it unusual to perform this poorly? We let X = number of missed free throws, so X ~ B(12,.25), and compute P ( X ≥ 5). Solution. We use the probability of the complement to obtain P ( X ≥ 5) = 1 – P ( X ≤ 4) , which is computed by 1-binomcdf(12,.25,4) ≈ 0.1576. If you are using a TI-89, simply enter a low end of 5 and a high end of 12.
Normal Approximations We conclude this section by showing how to approximate a binomial probability with a normal distribution. Example 13.7 Attitudes toward shopping. Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that in fact 60% of all adult U.S. residents would say “Agree” if asked the same question. What is the probability that 1520 or more of the sample agree? Solution. With n = 2500 and p = 0.60, we have both np ≥ 10 and n(1 – p) ≥10, so the distribution of X, the number who would agree is approximately normal with μ = np = 2500*0.6 = 1500 and
σ = np(1 − p) = 2500*.6*.4 = 24.495. . Thus, P ( X ≥ 1520) can be found (approximately) using the N(1500, 24.495) distribution. The command normalcdf(1520,1E99,1500,24.495) gives a probability of about 0.2071.
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We can find this probability exactly using binomcdf, remembering that at least 1520 is the complement of 1519 or less. The exact probability of 0.2131 is about 0.6% larger than the approximation.
13.2 Common Errors Err:Domain This error is normally caused in these types of problems by specifying a probability as a number greater than 1 (in percent possibly instead of a decimal) or a value for n or x which is not an integer. Reenter the command giving p in decimal form. Pressing N will return you to the input screen to correct the error. This will also occur in older TI-83 calculators if n is too large in a binomial calculation; in that case, you need to use the normal approximation. Probabilities larger than 1 It can’t. As we’ve said before, if it looks more than 1 on the first glance, check the right hand side. This value is 9.7x10-18 or seventeen zeros followed by the leading 9.
13.3 Selected Exercise Solutions 13.5 If the student catches 70% of errors, 30% will be missed, so if X = number of errors missed, X is Binomial (assuming independence), n = 10, p = 0.30. To find the probability of missing exactly 3 of the 10 errors, use binompdf from the DISTR menu. This is 0.2668 (about 26.7% chance of this happening). To find the probability of 3 or more, we subtract the probability of 2 or less from 1. The probability of 2 or less is found using binomcdf. The chance of missing at least 3 errors is 0.6172. 13.11
The mean is μ = np = 1535*.27 = 414.45. The
standard deviation is σ = np (1 − p ) = 1535*.27 *.73 = 17.394. Use the Normal approximation (with normalcdf) to find the approximate probability of more than 415 is 48.7%. Using the binomial calculation, the probability of more than 415 is found as the complement of 415 or less.
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That probability is 47.4%. The two differ by 1.3%; the approximation is pretty good. Continue your practice with binomial distributions by trying the following exercises: 13.25 Random stock prices. 13.27 The pill. 13.29 The pill, continued. 13.31 Genetics. 13.33 High school equivalency. 13.35 Multiple-choice tests. 13.39 A whooping cough outbreak. 13.41 A mixed group: probabilities.