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.