《商务与经济统计习题答案（第8版中文版）SBE8-SM06.doc》由会员分享，可在线阅读，更多相关《商务与经济统计习题答案（第8版中文版）SBE8-SM06.doc（19页珍藏版）》请在三一文库上搜索。
1、Continuous Probability DistributionsChapter 6Continuous Probability DistributionsLearning Objectives1.Understand the difference between how probabilities are computed for discrete and continuous random variables.2.Know how to compute probability values for a continuous uniform probability distributi
2、on and be able to compute the expected value and variance for such a distribution.3.Be able to compute probabilities using a normal probability distribution. Understand the role of the standard normal distribution in this process.4.Be able to compute probabilities using an exponential probability di
3、stribution.5.Understand the relationship between the Poisson and exponential probability distributions.Solutions:1.a.b.P(x = 1.25) = 0. The probability of any single point is zero since the area under the curve above any single point is zero.c.P(1.0 x 1.25) = 2(.25) = .50d.P(1.20 x 1.5) = 2(.30) = .
4、602.a.b.P(x 135)= (1/20) (140 - 135) = 0.25 d.minutes4.a.b.P(.25 x .60) = 1 (.40) = .405.a.Length of Interval = 261.2 - 238.9 = 22.3b.Note: 1 / 22.3 = 0.045P(x 250) = (0.045)(250 - 238.9) = 0.4995 Almost half drive the ball less than 250 yards.c.P(x 255) = (0.045)(261.2 - 255) = 0.279d.P(245 x 260)
5、= (0.045)(260 - 245) = 0.675e.P(x 250) = 1 - P(x 250) = 1 - 0.4995 = 0.5005The probability of anyone driving it 250 yards or more is 0.5005. With 60 players, the expected number driving it 250 yards or more is (60)(0.5005) = 30.03. Rounding, I would expect 30 of these women to drive the ball 250 yar
6、ds or more. 6.a.P(12 x 12.05) = .05(8) = .40b.P(x 12.02) = .08(8) = .64c.Therefore, the probability is .04 + .64 = .687.a.P(10,000 x 12,000) = 2000 (1 / 5000) = .40The probability your competitor will bid lower than you, and you get the bid, is .40.b.P(10,000 x 14,000) = 4000 (1 / 5000) = .80c.A bid
7、 of $15,000 gives a probability of 1 of getting the property.d.Yes, the bid that maximizes expected profit is $13,000.The probability of getting the property with a bid of $13,000 isP(10,000 x 72) = 1 - 0.8413 = 0.1587b.At x = 60P(x 60) = 0.5000 + 0.4986 = 0.9986 P(x 72) = 1 - 0.1587 = 0.841318.a.Fi
8、nd P(x 60)At x = 60P(x 60) = 0.5000 + 0.2549 = 0.7549 P(x 60) = 1 - P(x 5.0) = P(z 1.88) = 1 - P(z 1.88) = 1 - .9699 = .0301The rainfall exceeds 5 inches in 3.01% of the Aprils.b.P(x 3.0) = P(z .63) = 1 - P(z 40) = P(z 1.63) = 1 - P(z 1.63) = 1 - .9484 = .05165.16% of subscribers spend over 40 hours
9、 per week using the computer.c.A z-value of .84 cuts off an area of .20 in the upper tail. x = 27 + .84(8) = 33.72A subscriber who uses the computer 33.72 hours or more would be classified as a heavy user.21.From the normal probability tables, a z-value of 2.05 cuts off an area of approximately .02
10、in the upper tail of the distribution.x = m + zs = 100 + 2.05(15) = 130.75A score of 131 or better should qualify a person for membership in Mensa.22. Use m = 441.84 and s = 90a.At 400At 500P(0 z .65) = .2422 P(-.46 z 100)At x = 100P(x 100) = P(z .5) = 0.6915b.Find P(x 90)At x = 90P(x 90) = .5000 -
11、.3413 = 0.1587c.Find P(80 x 130)At x = 130P(x 130) = 0.8413At x = 80 Area to left is .0668P(80 x 130) = .8413 - .0668 = .774526.a.P(x 6) = 1 - e-6/8 = 1 - .4724 = .5276b.P(x 4) = 1 - e-4/8 = 1 - .6065 = .3935c.P(x 6) = 1 - P(x 6) = 1 - .5276 = .4724d.P(4 x 6) = P(x 6) - P(x 4) = .5276 - .3935 = .134
12、127.a.b.P(x 2) = 1 - e-2/3 = 1 - .5134 = .4866c.P(x 3) = 1 - P(x 3) = 1 - (1 - ) = e-1 = .3679d.P(x 5) = 1 - e-5/3 = 1 - .1889 = .8111e.P(2 x 5) = P(x 5) - P(x 2)= .8111 - .4866 = .324528.a.P(x 30) = 1 - P(x 30) = 1 - (1 - e-30/20 ) = e-30/20 = .2231c.P(10 x 30) = P(x 30) - P(x 10) = (1 - e-30/20 )
13、- (1 - e-10/20 ) = e-10/20 - e-30/20 = .6065 - .2231 = .383429.a.b.P(x 12) = 1 - e-12/12 = 1 - .3679 = .6321c.P(x 6) = 1 - e-6/12 = 1 - .6065 = .3935d.P(x 30)= 1 - P(x 30)= 1 - (1 - e-30/12)= .082130.a.50 hoursb.P(x 25) = 1 - e-25/50 = 1 - .6065 = .3935c.P(x 100)= 1 - (1 - e-100/50)= .135331.a.P(x 5
14、) = 1 - P(x 5) = 1 - (1 - e-5/2.78 ) = e-5/2.78 = .1655c.P(x 2.78) = 1 - P(x 2.78) = 1 - (1 - e-2.78/2.78 ) = e-1 = .3679This may seem surprising since the mean is 2.78 minutes. But, for the exponential distribution, the probability of a value greater than the mean is significantly less than the pro
15、bability of a value less than the mean.32.a.If the average number of transactions per year follows the Poisson distribution, the time between transactions follows the exponential distribution. So, m = of a year and then f(x) = 30 e-30xb.A month is 1/12 of a year so, The probability of no transaction
16、 during January is the same as the probability of no transaction during any month: .0821c.Since 1/2 month is 1/24 of a year, we compute,33.a.Let x = sales price ($1000s)b.P(x 215) = (1 / 25) (225 - 215) = 0.40c.P(x 80,000)At x = 80,000P(x 80,000) = 1.0000 - .8708 = 0.1292d.Find the z-score that cuts
17、 off 5% in the upper tail.z-score = 1.645. Solve for x. x = 63,000 + 1.645 (15,000)= 87,675The upper 5% of mortgage debt is in excess of $87,675.35.a.P(defect)= 1 - P(9.85 x 10.15)= 1 - P(-1 z 1)= 1 - .6826= .3174Expected number of defects = 1000(.3174) = 317.4b.P(defect)= 1 - P(9.85 x 10.15)= 1 - P
18、(-3 z 3)= 1 - .9972= .0028Expected number of defects = 1000(.0028) = 2.8c.Reducing the process standard deviation causes a substantial reduction in the number of defects.36.a.At 11%, z = -1.23b. Area to left is .5000 - .3255 = .3745 Area to left is .9744P (2000 x 2500) = .9744 - .3745 = .5999c.z = -
19、1.88x = 2071 - 1.88 (220.33) = $1656.7837.m = 10,000 s = 1500a.At x = 12,000 Area to left is .9082P(x 12,000) = 1.0000 - .9082 = .0918b.At .95Therefore, x = 10,000 + 1.645(1500) = 12,468.12,468 tubes should be produced.38.a.At x = 200 Area = .4772P(x 200) = .5 - .4772 = .0228b.Expected Profit= Expec
20、ted Revenue - Expected Cost= 200 - 150 = $5039.a.Find P(80,000 x 150,000)At x = 150,000P(x 150,000) = 0.7823At x = 80,000P(x 80,000) = .5000 - .4406 = 0.0594P(80,000 x 150,000) = 0.7823 - 0.0594 = 0.7229b.Find P(x 50,000)At x = 50,000P(x 50,000) = .5000 - .4948 = 0.0052c.Find the z-score cutting off 95% in the left tail.z-score = 1.645. Solve for x. x = 126,681 + 1.645 (30,000)= 176,031The probability is 0.95 that the number of lost jobs will not exceed 176,031.40.a.At 400,Area to left is .3085At 500,Area to left is .6915