11 a 1 2 209 209 0 d p p p µ µ b 2 2 2 2 1 2 48 48

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11.(a)1220.920.90DPPPµµ.(b)2222124.84.86.788DVVV´´.(c)±²±²6 or620.880.3789.DDP xxP z!³ µ!12.(a)Geometric probability:±² ±²40.840.160.0797(b) Binomial probability:±²±²5binomcdf 10,0.84,50.0130P Xd.(c)From part (b), f the soccer player’s success rate were still 84%, the probability that he wouldscore 5 or few goals in 10 penalty kicks is 0.0130.This is so low that we should be suspiciousabout whether he can still hit 84% of his shots.We have convincing evidence that his penaltykick success rate has fallen below 84%.
294The Practice of Statistics, 4/e- Chapter 6© 2011 BFW PublishersTest 6BPart 11.e±²21(0)(1)0.80P XPPtµµ2.a320.40.11, so 50.5 and0.10CCCC´´´3.b±²±²±²±²±²±²±²20,0000.605,0000.2512,0000.15$11,450XE xP´´ µ4.dBinomial probability formula:P(ksuccesses inntrials whenp= success in one trial) is±²1nkknppkµ§·µ¨¸©¹.5.dWhen adding or subtracting two independent variables, variances are added.6.a±²±²3252503251.50.066850P xPzP zµ§·!!!¨¸©¹7.eSince we are sampling without replacement, draws from the deck are not independent,violating one property of both binomial and geometric distributions.8.d±²±²180.91.62XnpP;±²±²±²±²1180.90.11.273XnppVµ9.aStatements II and III are conditions for the geometric setting.(Statement II is only truefor the binomial setting).Part 210.(a)± ²±²± ²±²± ²±²±²±²10.220.540.2100.13XP´´´; If this kind of investment was mademany times, the average profit would be 3 million dollars.(b)±²±²±²±²222220.2 130.5 230.2 430.1 1036.4XVµ´µ´µ´µ,6.42.53;XVTheexpected average distance between the amount of profit and the mean profit, if this kind ofinvestment were made many times, is about 2.53 million dollars.(c)± ²0.9 30.2$2.50YPµ, and±²0.9 2.53$2.28YV.11.(a)122662660DPPPµµ.(b)222212161622.63DVVV´´.(c)±²±²25 or2521.1050.2692.DDP xxP z!³ µ!12.(a)Geometric probability:±² ±²20.30.70.063(b)±²5binomcdf (10,0.7,5)0.1503P Xd(c)From part (b), if Pete’s success rate were still 70%, the probability that he would hit 5 orfewer bull’s eyes in 10 throw is 0.1503.While Pete didn’t do as well as we might expect, thisdeparture from the expected is not uncommon and could be attributed to chance variation.
© 2011 BFW PublishersThe Practice of Statistics, 4/e- Chapter 6295Test 6CPart 11.b±²± ²±²±²±²±²±²±²±²00.605000.0510000.1320000.22$595XE xP´´´2.dThe mean of a random variable is the long-run average value of that variable.We canexpect the cost per car to approach $595 as the number of cars insured gets very large.3.cVariables in I and III take on discrete integer values, the variable in II is continuous(though actual values are limited by the measurement process, there are no explicit gapsbetween values).4.cz-value for 0.60 is 0.25.±²±²$2400.2560$255´5.cWhen subtracting two independent variables, variances are added.
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