MAT 155. Key Concept. 155S5.3_3 Binomial Probability Distributions. September 21, Chapter 5 Probability Distributions

Transcription

1 MAT 155 Dr. Claude Moore Cape Fear Community College Chapter 5 Probability Distributions 5 1 Review and Preview 5 2 Random Variables 5 3 Binomial Probability Distributions 5 4 Mean, Variance, and Standard Deviation for the Binomial Distribution 5 5 Poisson Probability Distributions Find the Excel program in Important Links webpage, Technology, Mathematical Modeling & Statistics: Binomial Distribution (xls) Key Concept This section presents a basic definition of a binomial distribution along with notation, and methods for finding probability values. Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died. This video is a good introduction to Binomial Probability. distribution 1?playlist=Probability Binomial Probability Distribution A binomial probability distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure). 4. The probability of a success remains the same in all trials. Notation for Binomial Probability Distributions S and F (success and failure) denote the two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so P(S) = p P(F) = 1 p = q (p = probability of success) (q = probability of failure) 1

2 Notation (continued) n denotes the fixed number of trials. x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. p denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. P(x) denotes the probability of getting exactly x successes among the n trials. Methods for Finding Probabilities We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution. 1. Formula 2. Technology 3. Table A 1 Method 1: Using the Binomial Probability Formula Method 2: Using Technology STATDISK, Minitab, Excel, SPSS, SAS and the TI 83/84 Plus calculator can be used to find binomial probabilities. STATDISK MINITAB where n = number of trials x = number of successes among n trials p = probability of success in any one trial q = probability of failure in any one trial (q = 1 p) 2

3 Method 2: Using Technology continued STATDISK, Minitab, Excel and the TI 83 Plus calculator can all be used to find binomial probabilities. EXCEL TI 83 PLUS Calculator Method 3: Using Table A 1 in Appendix A Part of Table A 1 is shown below. With n = 12 and p = 0.80 in the binomial distribution, the probabilities of 4, 5, 6, and 7 successes are 0.001, 0.003, 0.016, and respectively. Find the Excel program in Important Links webpage, Technology, Mathematical Modeling & Statistics: Binomial Distribution (xls) Find the TI program in Important Links webpage, TI Calculator, Distributions (2ND VARS) binompdf Rationale for the Binomial Probability Formula Strategy for Finding Binomial Probabilities 1. Use computer software or a TI 83 Plus calculator if available. 2. If neither software nor the TI 83 Plus calculator is available, use Table A 1, if possible. 3. If neither software nor the TI 83 Plus calculator is available and the probabilities can t be found using Table A 1, use the binomial probability formula. 3

4 Important Hints Be sure that x and p both refer to the same category being called a success. When sampling without replacement, consider events to be independent if n < 0.05N. Recap In this section we have discussed: The definition of the binomial probability distribution. Notation. Important hints. Three computational methods. Rationale for the formula. In Exercises 5 12, determine whether or not the given procedure results in a binomial distribution. For those that are not binomial, identify at least one requirement that is not satisfied. 231/6. Clinical Trial of Lipitor Treating 863 subjects with Lipitor (Atorvastatin) and asking each subject How does your head feel? (based on data from Pfizer, Inc.). 231/8. Gender Selection Treating 152 couples with the YSORT gender selection method developed by the Genetics & IVF Institute and recording the gender of each of the 152 babies that are born. In Exercises 5 12, determine whether or not the given procedure results in a binomial distribution. For those that are not binomial, identify at least one requirement that is not satisfied. 231/10. Surveying Governors Fifteen different Governors are randomly selected from the 50 Governors currently in office and the sex of each Governor is recorded. NOTE: After thought, I changed my answer. No. Requirements 2 and 4 are not met. Since 15/50 = 0.30 > 0.05, and the selections are done without replacement, the trials cannot be considered independent. The value p of obtaining a success changes from trial to trial as each selection without replacement changes the population from which the next selection is made. 231/12. Surveying Statistics Students Two hundred statistics students are randomly selected and each is asked if he or she owns a TI 84 Plus calculator. 4

5 In Exercises 15 20, assume that a procedure yields a binomial distribution with a trial repeated n times. Use Table A 1 to find the probability of x successes given the probability p of success on a given trial. 232/16. n = 5, x = 1, p = 0.95 In Exercises 15 20, assume that a procedure yields a binomial distribution with a trial repeated n times. Use Table A 1 to find the probability of x successes given the probability p of success on a given trial. 232/20. n = 12, x = 12, p = 0.70 In Exercises 21 24, assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. 232/21. n = 12, x = 10, p = 3/4 In Exercises 21 24, assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. 232/24. n = 15, x = 13, p = 1/3 5

6 In Exercises 25 28, refer to the accompanying display. (When blood donors were randomly selected, 45% of them had blood that is Group O (based on data from the Greater New York Blood Pro gram).) The display shows the probabilities obtained by entering the values of n = 5 and p = /26. Group O Blood Find the probability that at least 3 of the 5 donors have Group O blood. If at least 3 Group O donors are needed, is it very likely that at least 3 will be obtained? In Exercises 25 28, refer to the accompanying display. (When blood donors were randomly selected, 45% of them had blood that is Group O (based on data from the Greater New York Blood Pro gram).) The display shows the probabilities obtained by entering the values of n = 5 and p = /28. Group O Blood Find the probability that at most 2 of the 5 donors have Group O blood. 233/34. Genetics Ten peas are generated from parents having the green yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 10 offspring peas, at least 1 has a green pod. Why does the usual rule for rounding (with three significant digits) not work in this case? 233/36. Genetics Slot Machine The author purchased a slot machine configured so that there is a probability of winning the jackpot on any individual trial. Although no one would seriously consider tricking the author, suppose that a guest claims that she played the slot machine 5 times and hit the jackpot twice. a. Find the probability of exactly 2 jackpots in 5 trials. b. Find the probability of at least 2 jackpots in 5 trials. c. Does the guest s claim of hitting 2 jackpots in 5 trials seem valid? Explain. 6

7 233/40. Job Interview Survey In a survey of 150 senior executives, 47% said that the most common job interview mistake is to have little or no knowledge of the company. a. If 6 of those surveyed executives are randomly selected without replacement for a follow up survey, find the probability that 3 of them said that the most common job interview mistake is to have little or no knowledge of the company. b. If part (a) is changed so that 9 of the surveyed executives are to be randomly selected with out replacement, explain why the binomial probability formula cannot be used. 233/44. Improving Quality The Write Right Company manufactures ballpoint pens and has been experiencing a 6% rate of defective pens. Modifications are made to the manufacturing process in an attempt to improve quality. The manager claims that the modified procedure is better because a test of 60 pens shows that only 1 is defective. a. Assuming that the 6% rate of defects has not changed, find the probability that among 60 pens, exactly 1 is defective. b. Assuming that the 6% rate of defects has not changed, find the probability that among 60 pens, none are defective. c. What probability value should be used for determining whether the modified process results in a defect rate that is less than 6%? d. What can you conclude about the effectiveness of the modified manufacturing process? Let x = the number who said the most common mistake is not to know the company. binomial problem: n = 6 and p = 0.47 a. P x=3 = binompdf 6,0.47,3 = b. The binomial distribution requires that the repeated selections be independent. Since these persons are selected from the original group of 150 without replacement, the repeated selections are not independent and the binomial distribution should not be used. In part a, however, the sample size is 6/150 = 4.0% 5% of the population and the repeated samples may be treated as though they are independent. If the sample size is increased to 9, the sample is 9/150 = 6.0% > 5% of the population and the criteria for using independence to get an approximate probability is no longer met. Let x = the number of defective pens if there has been no change in the rate. binomial problem: n = 60 and p = 0.06 a. P x=1 = binompdf 60,0.06,1 = b. P x=0 = binompdf 60,0.06,0 = c. In situations involving multiple ordered outcomes, the unusualness of a particular outcome is generally determined by the probability of getting that outcome or a more extreme outcome. In this instance the appropriate probability is P x 1 = P x=0 + P x=1 = = , rounded to d. Since > 0.05, the observed outcome would not be considered unusual under the old rate. There is not sufficient evidence to conclude that the modified procedure is better. 7