Chapter 5 Black 8e Student Solutions Manual Ch05
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Student’s Solutions Manual and Study Guide: Chapter 5 Page 1 Chapter 5 Discrete Distributions LEARNING OBJECTIVES The overall learning objective of Chapter 5 is to help you understand a category of probability distributions that produces only discrete outcomes, thereby enabling you to: 1. 2. 3. 4. 5. Define a random variable in order to differentiate between a discrete distribution and a continuous distribution Determine the mean, variance, and standard deviation of a discrete distribution Solve problems involving the binomial distribution using the binomial formula and the binomial table Solve problems involving the Poisson distribution using the Poisson formula and the Poisson table Solve problems involving the hypergeometric distribution using the hypergeometric formula Student’s Solutions Manual and Study Guide: Chapter 5 Page 2 CHAPTER OUTLINE 5.1 Discrete Versus Continuous Distributions 5.2 Describing a Discrete Distribution Mean, Variance, and Standard Deviation of Discrete Distributions Mean or Expected Value Variance and Standard Deviation of a Discrete Distribution 5.3 Binomial Distribution Solving a Binomial Problem Using the Binomial Table Using the Computer to Produce a Binomial Distribution Mean and Standard Deviation of the Binomial Distribution Graphing Binomial Distributions 5.4 Poisson Distribution Working Poisson Problems by Formula Using the Poisson Tables Mean and Standard Deviation of a Poisson Distribution Graphing Poisson Distributions Using the Computer to Generate Poisson Distributions Approximating Binomial Problems by the Poisson Distribution 5.5 Hypergeometric Distribution Using the Computer to Solve for Hypergeometric Distribution Probabilities KEY TERMS Binomial Distribution Continuous Distributions Continuous Random Variables Discrete Distributions Discrete Random Variables Hypergeometric Distribution Lambda () Mean, or Expected Value Poisson Distribution Random Variable Student’s Solutions Manual and Study Guide: Chapter 5 Page 3 STUDY QUESTIONS 1. Variables that take on values at every point over a given interval are called _______________ _______________ variables. 2. If the set of all possible values of a variable is at most finite or a countably infinite number of possible values, then the variable is called a _______________ _______________ variable. 3. An experiment in which a die is rolled six times will likely produce values of a _______________ random variable. 4. An experiment in which a researcher counts the number of customers arriving at a supermarket checkout counter every two minutes produces values of a _______________ random variable. 5. An experiment in which the time it takes to assemble a product is measured is likely to produce values of a _______________ random variable. 6. A binomial distribution is an example of a _______________ distribution. 7. The normal distribution is an example of a _______________ distribution. 8. The long-run average of a discrete distribution is called the __________________ or ______________________________________. Use the following discrete distribution to answer 9 and 10 x P(x) 1 2 3 4 .435 .241 .216 .108 9. The mean of the discrete distribution above is ____________. 10. The variance of the discrete distribution above is _______________. 11. On any one trial of a binomial experiment, there can be only _______________ possible outcomes. 12. Suppose the probability that a given part is defective is .10. If four such parts are randomly drawn from a large population, the probability that exactly two parts are defective is ________. 13. Suppose the probability that a given part is defective is .04. If thirteen such parts are randomly drawn from a large population, the expected value or mean of the binomial distribution that describes this experiment is ________. 14. Suppose a binomial experiment is conducted by randomly selecting 20 items where p = .30. The standard deviation of the binomial distribution is _______________. Student’s Solutions Manual and Study Guide: Chapter 5 Page 4 15. Suppose forty-seven percent of the workers in a large corporation are under thirty-five years of age. If fifteen workers are randomly selected from this corporation, the probability of selecting exactly ten who are under thirty-five years of age is _______________. 16. Suppose that twenty-three percent of all adult Americans fly at least once a year. If twelve adult Americans are randomly selected, the probability that exactly four have flown at least once last year is _______________. 17. Suppose that sixty percent of all voters support the President of the United States at this time. If twenty voters are randomly selected, the probability that at least eleven support the President is _______________. 18. The Poisson distribution was named after the French mathematician _______________. 19. The Poisson distribution focuses on the number of discrete occurrences per _______________. 20. The Poisson distribution tends to describe _______________ occurrences. 21. The long-run average or mean of a Poisson distribution is _______________. 22. The variance of a Poisson distribution is equal to _______________. 23. If Lambda is 2.6 occurrences over an interval of five minutes, the probability of getting six occurrences over one five minute interval is _______________. 24. Suppose that in the long-run a company determines that there are 1.2 flaws per every twenty pages of typing paper produced. If ten pages of typing paper are randomly selected, the probability that more than two flaws are found is _______________. 25. If Lambda is 1.8 for a four minute interval, an adjusted new Lambda of _______ would be used to analyze the number of occurrences for a twelve minute interval. 26. Suppose a binomial distribution problem has an n = 200 and a p = .03. If this problem is worked using the Poisson distribution, the value of Lambda is ________. 27. The hypergeometric distribution should be used when a binomial type experiment is being conducted without replacement and the sample size is greater than or equal to ________% of the population. 28. Suppose a population contains sixteen items of which seven are X and nine are Y. If a random sample of five of these population items is selected, the probability that exactly three of the five are X is ________. 29. Suppose a population contains twenty people of which eight are members of the Catholic church. If a sample of four of the population is taken, the probability that at least three of the four are members of the Catholic church is ________. 30. Suppose a lot of fifteen personal computer printers contains two defective printers. If three of the fifteen printers are randomly selected for testing, the probability that no defective printers are selected is _______________. Student’s Solutions Manual and Study Guide: Chapter 5 Page 5 ANSWERS TO STUDY QUESTIONS 1. Continuous Random 16. .1712 2. Discrete Random 17. .755 3. Discrete 18. Poisson 4. Discrete 19. Interval 5. Continuous 20. Rare 6. Discrete 21. Lambda 7. Continuous 22. Lambda 8. Mean, Expected Value 23. .0319 9. 1.997 24. .0232 10. 1.083 25. 5.4 11. Two 26. 6.0 12. .0486 27. 5 13. 0.52 28. .2885 14. 2.049 29. .1531 15. .0661 30. .6286 Student’s Solutions Manual and Study Guide: Chapter 5 Page 6 SOLUTIONS TO THE ODD-NUMBERED PROBLEMS IN CHAPTER 5 5.1 x 1 2 3 4 5 P(x) .238 .290 .177 .158 .137 µ = [x·P(x)] = 2.666 = 5.3 x 0 1 2 3 4 (x-µ)2 (x-µ)2·P(x) 2.775556 0.6605823 0.443556 0.1286312 0.111556 0.0197454 1.779556 0.2811700 5.447556 0.7463152 2 2 = [(x-µ) ·P(x)] = 1.836444 1.836444 = 1.355155 x·P(x) .238 .580 .531 .632 .685 P(x) x·P(x) (x-µ)2 (x-µ)2·P(x) .461 .000 0.913936 0.421324 .285 .285 0.001936 0.000552 .129 .258 1.089936 0.140602 .087 .261 4.177936 0.363480 .038 .152 9.265936 0.352106 2 2 E(x) = µ = [x·P(x)]= 0.956 = [(x-µ) ·P(x)] = 1.278064 = 1.278064 = 1.1305 Student’s Solutions Manual and Study Guide: Chapter 5 5.5 a) n=4 P(x=3) = b) n=7 p = .10 3 1 4C3(.10) (.90) p = .80 Page 7 q = .90 = 4(.001)(.90) = .0036 q = .20 P(x=4) = 7C4(.80)4(.20)3 = 35(.4096)(.008) = .1147 c) n = 10 p = .60 q = .40 P(x > 7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) = 7 3 8 2 9 1 10 0 10C7(.60) (.40) + 10C8(.60) (.40) + 10C9(.60) (.40) +10C10(.60) (.40) 120(.0280)(.064) + 45(.0168)(.16) + 10(.0101)(.40) + 1(.0060)(1) = .2150 + .1209 + .0403 + .0060 = .3822 d) n = 12 p = .45 q = .55 P(5 < x < 7) = P(x=5) + P(x=6) + P(x=7) = 5 7 6 6 7 5 12C5(.45) (.55) + 12C6(.45) (.55) + 12C7(.45) (.55) = 792(.0185)(.0152) + 924(.0083)(.0277) + 792(.0037)(.0503) = .2225 + .2124 + .1489 = .5838 = Student’s Solutions Manual and Study Guide: Chapter 5 5.7 a) n = 20 p = .70 Page 8 q = .30 µ = np = 20(.70) = 14 = b) n p q 20(.70)(.30) 4.2 = 2.05 n = 70 p = .35 q = .65 µ = np = 70(.35) = 24.5 = c) n p q 70(.35)(.65) 15.925 = 3.99 n = 100 p = .50 q = .50 µ = np = 100(.50) = 50 = n p q 100(.50)(.50) 25 = 5 5.9 Looking at the graph, the highest probability appears to be at x = 1 followed by x = 2. It is most likely that the mean falls between x = 1 and x = 2 but because of the size of the drop from x = 1 to x = 2, the mean is much closer to x =1. Thus, the mean is something like 1.2. Since the expected value = n∙ p, if n = 6 and is 1.2, solving for p results in p = .20. So, p is near to .20 and the mean or expected value is around 1.2. 5.11 a) n = 20 20C8 b) c) x=8 (.27)8(.73)12 = 125,970(.000028243)(.022902) = .0815 n = 20 20C0 p = .27 p = .30 x=0 (.30) 0(.70)20 = (1)(1)(.0007979) = .0007979 n = 20 p = .30 x>7 Use table A.2: P(x=8) + P(x=9) + . . . + P(x=14)= .114 + .065 + .031 + .012 + .004 + .001 + .000 + … = .227 Student’s Solutions Manual and Study Guide: Chapter 5 5.13 n = 25 Page 9 p = .60 a) x > 15 P(x > 15) = P(x = 15) + P(x = 16) + · · · + P(x = 25) Using Table A.2 x 15 16 17 18 19 20 21 22 n = 25, p = .60 Prob .161 .151 .120 .080 .044 .020 .007 .002 .585 b) x > 20 P(x > 20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25) = Using Table A.2 n = 25, p = .60 .007 + .002 + .000 + .000 + .000 = .009 c) P(x < 10) Using Table A.2 x 9 8 7 <6 n = 25, p = .60 and x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Prob. .009 .003 .001 .000 .013 Student’s Solutions Manual and Study Guide: Chapter 5 5.15 n = 15 a) P(x = 5) = Page 10 p = .20 5 10 15C5(.20) (.80) = 3003(.00032)(.1073742) = .1032 b) P(x > 9): Using Table A.2 P(x = 10) + P(x = 11) + . . . + P(x = 15) = .000 + .000 + . . . + .000 = .000 c) P(x = 0) = 0 15 15C0(.20) (.80) = (1)(1)(.035184) = .0352 d) P(4 < x < 7): Using Table A.2 P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) = .188 + .103 + .043 + .014 = .348 e) Graph Student’s Solutions Manual and Study Guide: Chapter 5 5.17 a) P(x=5 = 2.3) = Page 11 2.35 e 2.3 (64.36343)(.100259) = .0538 5! 120 3.9 2 e 3.9 (15.21)(.020242) b) P(x=2 = 3.9) = = .1539 2! 2 c) P(x < 3 = 4.1) = P(x=3) + P(x=2) + P(x=1) + P(x=0) = 4.13 e 4.1 (68.921)(.016573) = .1904 3! 6 4.12 e 4.1 (16.81)(.016573) = .1393 2! 2 4.11 e 4.1 (4.1)(.016573) = .0679 1! 1 4.10 e 4.1 (1)(.016573) = .0166 0! 1 .1904 + .1393 + .0679 + .0166 = .4142 d) P(x=0 = 2.7) = 2.7 0 e 2.7 (1)(.06721) = .0672 0! 1 e) P(x=1 = 5.4)= 5.41 e 5.4 (5.4)(.0045166) = .0244 1! 1 f) P(4 < x < 8 = 4.4): P(x=5 = 4.4) + P(x=6 = 4.4) + P(x=7 = 4.4)= 4.4 5 e 4.4 4.4 6 e 4.4 4.4 7 e 4.4 + + = 6! 7! 5! (1649.1622)(.01227734) (7256.3139)(.01227734) (31,927.781)(.01227734) + + 120 720 5040 = .1687 + .1237 + .0778 = .3702 Student’s Solutions Manual and Study Guide: Chapter 5 5.19 a) = 6.3 mean = 6.3 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Page 12 Standard deviation = Prob .0018 .0116 .0364 .0765 .1205 .1519 .1595 .1435 .1130 .0791 .0498 .0285 .0150 .0073 .0033 .0014 .0005 .0002 .0001 .0000 6.3 = 2.51 Student’s Solutions Manual and Study Guide: Chapter 5 b) = 1.3 mean = 1.3 x 0 1 2 3 4 5 6 7 8 9 Page 13 standard deviation = Prob .2725 .3542 .2303 .0998 .0324 .0084 .0018 .0003 .0001 .0000 1.3 = 1.14 Student’s Solutions Manual and Study Guide: Chapter 5 c) = 8.9 mean = 8.9 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Page 14 standard deviation = Prob .0001 .0012 .0054 .0160 .0357 .0635 .0941 .1197 .1332 .1317 .1172 .0948 .0703 .0481 .0306 .0182 .0101 .0053 .0026 .0012 .0005 .0002 .0001 8.9 = 2.98 Student’s Solutions Manual and Study Guide: Chapter 5 d) = 0.6 mean = 0.6 x 0 1 2 3 4 5 6 Page 15 standard deviation = Prob .5488 .3293 .0988 .0198 .0030 .0004 .0000 0.6 = .775 Student’s Solutions Manual and Study Guide: Chapter 5 5.21 Page 16 = x/n = 126/36 = 3.5 Using Table A.3 a) P(x = 0) = .0302 b) P(x > 6) = P(x = 6) + P(x = 7) + . . . = .0771 + .0385 + .0169 + .0066 + .0023 + .0007 + .0002 + .0001 = .1424 c) P(x < 4 10 minutes) Double Lambda to = 7.010 minutes P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = .0009 + .0064 + .0223 + .0521 = .0817 d) P(3 < x < 6 10 minutes) = 7.0 10 minutes P(3 < x < 6) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = .0521 + .0912 + .1277 + .1490 = .42 e) P(x = 8 15 minutes) Change Lambda for a 15 minute interval by multiplying the original Lambda by 3. = 10.5 15 minutes P(x = 815 minutes) = x e x! (10.58 )(e 10.5 ) 8! = .1009 Student’s Solutions Manual and Study Guide: Chapter 5 5.23 Page 17 = 1.2 collisions4 months a) P(x=0 = 1.2): from Table A.3 = .3012 b) P(x=22 months): The interval has been decreased (by ½) New Lambda = = 0.6 collisions2 months P(x=2 = 0.6): from Table A.3 = .0988 c) P(x < 1 collision6 months): The interval length has been increased (by 1.5) New Lambda = = 1.8 collisions6 months P(x < 1 = 1.8): from Table A.3 x 0 1 Prob. .1653 .2975 .4628 The result is likely to happen almost half the time (46.26%). Ship channel and weather conditions are about normal for this period. Safety awareness is about normal for this period. There is no compelling reason to reject the lambda value of 0.6 collisions per 4 months based on an outcome of 0 or 1 collisions per 6 months. Student’s Solutions Manual and Study Guide: Chapter 5 5.25 n = 100,000 Page 18 p = .00004 a) P(x > 7n = 100,000 p = .00004): = µ = np = 100,000(.00004) = 4.0 Since n > 20 and np < 7, the Poisson approximation to this binomial problem is close enough. P(x > 7 = 4): Using Table A.3 x 7 8 9 10 11 12 13 14 Prob. .0595 .0298 .0132 .0053 .0019 .0006 .0002 .0001 .1106 x 11 12 13 14 Prob. .0019 .0006 .0002 .0001 .0028 b) P(x >10 = 4): Using Table A.3 c) Since getting more than 10 is a rare occurrence, this particular geographic region appears to have a higher average rate than other regions. An investigation of particular characteristics of this region might be warranted. Student’s Solutions Manual and Study Guide: Chapter 5 5.27 Page 19 a) P(x = 3 N = 11, A = 8, n = 4) 8 C3 3 C1 (56)(3) = .5091 330 11 C 4 b) P(x < 2)N = 15, A = 5, n = 6) P(x = 1) + P (x = 0) = 5 C1 10 C5 C C (5)(252) (1)(210) + 5 0 10 6 = 5005 5005 15 C 6 15 C 6 .2517 + .0420 = .2937 c) P(x=0 N = 9, A = 2, n = 3) 2 C0 7 C3 (1)(35) = .4167 84 9 C3 d) P(x > 4 N = 20, A = 5, n = 7) = P(x = 5) + P(x = 6) + P(x = 7) = 5 C5 15 C 2 + 20 C 7 5 C 6 15 C1 + 20 C 7 5 C 7 15 C 0 = 20 C 7 (1)(105) + 5C6 (impossible) + 5C7(impossible) = .0014 77520 Student’s Solutions Manual and Study Guide: Chapter 5 5.29 N = 17 8 a) P(x = 0) = 8 b) P(x = 4) = Page 20 A=8 C 0 9 C 4 (1)(126) = = .0529 2380 17 C 4 C 4 9 C 0 (70)(1) = = .0294 2380 17 C 4 c) P(x = 2 non computer) = 5.31 N = 10 C 2 8 C 2 (36)(28) = = .4235 2380 17 C 4 P(x = 2): C2 8 C2 (1)(28) .1333 C 210 10 4 b) A = 5 x = 0 5 P(x = 0): C0 5 C4 (1)(5) .0238 C 210 10 4 c) A = 4 x = 3 4 5.33 9 n=4 a) A = 2 x = 2 2 n=4 P(x = 3): C3 6 C1 (4)(6) .1143 C4 210 10 N = 18 A = 11 Hispanic n=5 P(x < 1) = P(1) + P(0) = 11 C1 7 C 4 + 18 C 5 11 C 0 7 C5 (11)(35) (1)(21) = = .0449 + .0025 = .0474 8568 8568 18 C 5 It is fairly unlikely that these results occur by chance. A researcher might want to further investigate this result to determine causes. Were officers selected based on leadership, years of service, dedication, prejudice, or some other reason? Student’s Solutions Manual and Study Guide: Chapter 5 5.35 a) P(x = 14 n = 20 and p = .60) = .124 b) P(x < 5 n = 10 and p =.30) = P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x=0) = x 0 1 2 3 4 Prob. .028 .121 .233 .267 .200 .849 c) P(x > 12 n = 15 and p = .60) = P(x = 12) + P(x = 13) + P(x = 14) + P(x = 15) x 12 13 14 15 Prob. .063 .022 .005 .000 .090 d) P(x > 20 n = 25 and p = .40) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x=25) = x 21 22 23 24 25 Prob. .000 .000 .000 .000 .000 .000 Page 21 Student’s Solutions Manual and Study Guide: Chapter 5 5.37 a) P(x = 3 = 1.8) = .1607 b) P(x < 5 = 3.3) = P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) = x 0 1 2 3 4 Prob. .0369 .1217 .2008 .2209 .1823 .7626 c) P(x > 3 = 2.1) = x 3 4 5 6 7 8 9 10 11 Prob. .1890 .0992 .0417 .0146 .0044 .0011 .0003 .0001 .0000 .3504 d) P(2 < x < 5 = 4.2): P(x=3) + P(x=4) + P(x=5) = x 3 4 5 Prob. .1852 .1944 .1633 .5429 Page 22 Student’s Solutions Manual and Study Guide: Chapter 5 5.39 Page 23 n = 25 p = .20 retired a) from Table A.2: P(x = 7) = .1108 b) P(x > 10): P(x = 10) + P(x = 11) + . . . + P(x = 25) = .012 + .004 + .001 = .017 c) Expected Value = µ = np = 25(.20) = 5 d) n = 20 p = .40 mutual funds P(x = 8) = .1797 e) P(x < 6) = P(x = 0) + P(x = 1) + . . . + P(x = 5) = .000 + .000 + .003 +.012 + .035 + .075 = .125 f) P(x = 0) = .000 g) P(x > 12) = P(x = 12) + P(x = 13) + . . . + P(x = 20) = .035 + .015 + .005 + .001 = .056 h) x = 8 5.41 Expected Number = µ = n p = 20(.40) = 8 N = 32 A = 10 a) P(x = 3) = 10 b) P(x = 6) = 10 c) P(x = 0) = 10 d) A = 22 = n = 12 C3 22 C9 (120)(497,420) = = .2644 225,792,840 32 C12 C6 22 C 6 (210)(74,613) = = .0694 225,792,840 32 C12 C 0 22 C12 (1)(646,646) = = .0029 225,792,840 32 C12 P(7 < x < 9) = 22 C 7 10 C5 + 32 C12 22 C8 10 C 4 + 32 C12 (170,544)(252) (319,770)(210) (497,420)(120) 225,792,840 225,792,840 225,792,840 = .1903 + .2974 + .2644 = .7521 22 C9 10 C3 32 C12 Student’s Solutions Manual and Study Guide: Chapter 5 5.43 Page 24 a) n = 20 and p = .25 The expected number = µ = np = (20)(.25) = 5.00 b) P(x < 1 n = 20 and p = .25) = P(x = 1) + P(x = 0) = 1 19 20C1(.25) (.75) + 20C0(.25)0(.75)20 = (20)(.25)(.00423) + (1)(1)(.0032) = .0212 +. 0032 = .0244 Since the probability is so low, the population of your state may have a lower percentage of chronic heart conditions than those of other states. 5.45 n = 12 a.) P(x = 0 long hours): p = .20 0 12 12C0(.20) (.80) = .0687 b.) P(x > 6) long hours): p = .20 Using Table A.2: .016 + .003 + .001 = .020 c) P(x = 5 good financing): p = .25, 5 7 12C5(.25) (.75) = .1032 d.) p = .19 (good plan), expected number = µ = n(p) = 12(.19) = 2.28 Student’s Solutions Manual and Study Guide: Chapter 5 Page 25 P(x < 3) n = 8 and p = .60): 5.47 From Table A.2: x 0 1 2 3 Prob. .001 .008 .041 .124 .174 17.4% of the time in a sample of eight, three or fewer customers are walk-ins by chance. Other reasons for such a low number of walk-ins might be that she is retaining more old customers than before or perhaps a new competitor is attracting walk-ins away from her. = 1.2 hoursweek 5.49 a) P(x = 0 = 1.2) = (from Table A.3) .3012 b) P(x > 3 = 1.2) = (from Table A.3) x 3 4 5 6 7 8 Prob. .0867 .0260 .0062 .0012 .0002 .0000 .1203 a) P(x < 5 3 weeks) If = 1.2 for 1 week, the = 3.6 for 3 weeks. P(x < 5 = 3.6) = (from Table A.3) x 4 3 2 1 0 Prob. .1912 .2125 .1771 .0984 .0273 .7065 Student’s Solutions Manual and Study Guide: Chapter 5 5.51 N = 24 n=6 a) P(x = 6) = b) P(x = 0) = 8 Page 26 A=8 C6 16 C0 (28)(1) = .0002 134,596 24 C 6 8 C0 16 C6 (1)(8008) = .0595 134,596 24 C 6 d) A = 16 East Side P(x = 3) = 16 C3 8 C3 (560)(56) = .2330 C 134 , 596 24 6 Student’s Solutions Manual and Study Guide: Chapter 5 5.53 Page 27 = 2.4 calls1 minute a) P(x = 0 = 2.4) = (from Table A.3) .0907 b) Can handle x < 5 calls Cannot handle x > 5 calls P(x > 5 = 2.4) = (from Table A.3) x 6 7 8 9 10 11 c) P(x = 3 calls2 minutes) The interval has been increased 2 times. New Lambda: = 4.8 calls2 minutes. from Table A.3: .1517 d) P(x < 1 calls15 seconds): The interval has been decreased by ¼. New Lambda = = 0.6 calls15 seconds. P(x < 1 = 0.6) = (from Table A.3) P(x = 1) = .3293 P(x = 0) = .5488 .8781 Prob. .0241 .0083 .0025 .0007 .0002 .0000 .0358 Student’s Solutions Manual and Study Guide: Chapter 5 5.55 p = .005 Page 28 n = 1,000 = np = (1,000)(.005) = 5 a) P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = .0067 + .0337 + .0842 + .1404 = .265 b) P(x > 10) = P(x = 11) + P(x = 12) + . . . = .0082 + .0034 + .0013 + .0005 + .0002 = .0136 c) P(x = 0) = .0067 5.57 N = 25 a) n = 5 17 x=3 C3 8 C2 (680)(28) .3584 C 53 , 130 25 5 b) n = 8 x < 2 5 A = 17 A=5 C0 20 C8 5 C1 20 C7 5 C2 20 C6 C8 C8 C8 25 25 25 (1)(125,970) (5)(77,520) (10)(38,760) 1,081,575 1,081,575 1,081,575 .1165 + .3584 + .3584 = .8333 c) n = 5 x=2 A=3 2 3 5C2(3/25) (22/25) = (10)(.0144)(.681472) = .0981 Student’s Solutions Manual and Study Guide: Chapter 5 5.59 Page 29 a) = 3.051,000 for U.S. 3.050 e 3.05 P( x 0) .0474 0! b) = 6.102,000 for U.S. (just double previous lambda) 6.10 6 e 6.104 (51,520.37)(.002243) P( x 6) .1605 6! 720 c) = 1.071,000 and = 3.213,000 from Table A.3: P(x < 7) = P(x = 0) + P(x = 1) + . . . + P(x = 6) = (321 . 0 )(e 3.21 ) (321 . 1 )(e 3.21 ) (321 . 2 )(e 3.21 ) (321 . 3 )(e 3.21 ) (321 . 4 )(e 3.21 ) (321 . 5 )(e 3.21 ) (321 . 6 )(e 3.21 ) 0! 1! 2! 3! 4! 5! 6! .0404 + .1295 + .2079 + .2225 + .1785 + .1146 + .0613 = .9547 5.61 This printout contains the probabilities for various values of x from zero to eleven from a Poisson distribution with = 2.78. Note that the highest probabilities are at x = 2 and x = 3 which are near the mean. The probability is slightly higher at x = 2 than at x = 3 even though x = 3 is nearer to the mean because of the “piling up” effect of x = 0. 5.63 This is the graph of a Poisson Distribution with = 1.784. Note the high probabilities at x = 1 and x = 2 which are nearest to the mean. Note also that the probabilities for values of x > 8 are near to zero because they are so far away from the mean or expected value.
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