Then 18 d p and 2 2 31 40 5061 d v ² ² 0 18 0 36

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Then18DPand22314050.61.DV´±²±²0180.360.3594.50.6P DPzP zµ§·³³³ µ¨¸©¹Quiz 6.2B1.(a)±²±²0.10 68$6.80;0.10 9.5$0.95.DDPV(b)±²±²0.10 680.25 42$17.30;TP´±²±²220.10 9.50.25 5.8$1.73.TV´(c)$17.30$20$2.70GPµµ(a net loss for thecontestant);$1.73GV(not changed by subtracting a constant).2.LetD= difference in heightbetween random male and random female.Then5DPand2232.53.91DV´.±²±²0501.280.1003.3.91P DPzP zµ§·³³³ µ¨¸©¹
© 2011 BFW PublishersThe Practice of Statistics, 4/e- Chapter 6291Quiz 6.2C1.(a)±²±²22.5;.5 0.5.5 1.50.5HHPVµ´µ(b).5.5.5.52;HP´´´2222.5.5.5.51XV´´´.2.(a)±²±²1110010010022YP´;.±²±²22111.20.650.68222YV§·§·´¨¸¨¸©¹©¹(b)±²±²1210010010033YP´;±²±²22121.20.650.59033YV§·§·´¨¸¨¸©¹©¹.3.We are interested in the difference between their jobcompletion times, so105987min utes;DPµ22101518.03DV´.To be within 5minutes means ±5 minutes.Therefore±²±²575755.67.110.2048.18.0318.03PDPzPzµ µµ§·µ³³³³µ³³ µ¨¸©¹Quiz 6.3A1.(a) This is a geometric setting:Binaryoutcomes (HIV or not),Independenttrials (oneperson’s condition does not influence condition of next randomly-selected individual), we arecounting the number ofTrialsto the first HIV case, and the probability ofSuccess—finding aperson who is HIV-positive—is always 0.01.(b)This is neither binomial nor geometric,because each trial is not independent of previous trials since it is done without replacement: thesuit of the first card influences the probability of a heart on the second card, and so on.2.(a) Geometric setting withp= 0.24:±² ±²40.760.240.0801|.(b) Equivalent to±²±²4No successes in first 4 trials0.760.3336P; or±²±²±²5141geomcdf 0.24,40.3336.P xP xtµdµ3.(a)Fis binomial withn= 15 andp= 0.20.±²±² ±²3121530.20.80.25013P F§·¨¸©¹(b)±²±²±²43binomcdf 15,.2,30.6482P FP F³d(c)±²±²150.203FnpP,±²±²±²±²1150.20.81.55FnppVµ.Quiz 6.3B1.(a) This is neither binomial or geometric, because the probability of success (selecting afemale student) is probably different in each trial, unless every classroom has exactly the sameproportion of females, which is unlikely. (b) This is a geometric setting:Binaryoutcomes(more than two occupants or not),Independenttrials (the number of occupants of one car doesnot influence the number of occupants in the next randomly-selected car, we are counting thenumber ofTrialsto the first car with more than two occupants, and the probability ofSuccessfinding a car with more than two occupants—is always the same.2.(a)Xis binomial withn= 20 andp= 0.02.±²±² ±²1192010.020.980.27251P X§·¨¸©¹
292The Practice of Statistics, 4/e- Chapter 6© 2011 BFW Publishers(b)±²±²2binomcdf20,.02,20.9929P Xd(c)±²±²200.020.4;XnpP±²±²±²±²1200.020.980.63XnppVµ.

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Term
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Statistics, Standard Deviation, Probability theory, BFW Publishers, 4 e Chapter

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