商务与经济统计习题答案（第8版中文版）SBE8-SM06.doc

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 distribution 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 distribution.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) = .602.a.b.P(x < 15) = .10(5) = .50c.P(12 x 18) = .10(6) = .60d.e.3.a.b.P(x 130) = (1/20) (130 - 120) = 0.50c.P(x > 135)= (1/20) (140 - 135) = 0.25 d.minutes4.a.b.P(.25 < x < .75) = 1 (.50) = .50c.P(x .30)= 1 (.30) = .30d.P(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) = (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 yards 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 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 < 13,000) = 3000 (1 / 5000) = .60.The probability of not getting the property with a bid of $13,000 is .40.The profit you will make if you get the property with a bid of $13,000 is $3000 = $16,000 - 13,000. So your expected profit with a bid of $13,000 isEP ($13,000) = .6 ($3000) + .4 (0) = $1800.If you bid $15,000 the probability of getting the bid is 1, but the profit if you do get the bid is only $1000 = $16,000 - 15,000. So your expected profit with a bid of $15,000 isEP ($15,000) = 1 ($1000) + 0 (0) = $1,000.8.9.a.b.6826 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50.c.9544 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50.10.a.3413b.4332c.4772d.493811.a.3413These probability values are read directlyfrom the table of areas for the standard b.4332normal probability distribution. SeeTable 1 in Appendix B.c.4772d.4938e.498612.a.2967b.4418c.5000 - .1700 = .3300d.0910 + .5000 = .5910e.3849 + .5000 = .8849f.5000 - .2612 = .238813.a.6879 - .0239 = .6640b.8888 - .6985 = .1903c.9599 - .8508 = .109114.a.Using the table of areas for the standard normal probability distribution, the area of .4750 corresponds to z = 1.96.b.Using the table, the area of .2291 corresponds to z = .61.c.Look in the table for an area of .5000 - .1314 = .3686. This provides z = 1.12.d.Look in the table for an area of .6700 - .5000 = .1700. This provides z = .44.15.a.Look in the table for an area of .5000 - .2119 = .2881. Since the value we are seeking is below the mean, the z value must be negative. Thus, for an area of .2881, z = -.80.b.Look in the table for an area of .9030 / 2 = .4515; z = 1.66.c.Look in the table for an area of .2052 / 2 = .1026; z = .26.d.Look in the table for an area of .4948; z = 2.56.e.Look in the table for an area of .1915. Since the value we are seeking is below the mean, the z value must be negative. Thus, z = -.50.16.a.Look in the table for an area of .5000 - .0100 = .4900. The area value in the table closest to .4900 provides the value z = 2.33.b.Look in the table for an area of .5000 - .0250 = .4750. This corresponds to z = 1.96.c.Look in the table for an area of .5000 - .0500 = .4500. Since .4500 is exactly halfway between .4495 (z = 1.64) and .4505 (z = 1.65), we select z = 1.645. However, z = 1.64 or z = 1.65 are also acceptable answers.d.Look in the table for an area of .5000 - .1000 = .4000. The area value in the table closest to .4000 provides the value z = 1.28.17.Convert mean to inches: m = 69a.At x = 72P(x 72) = 0.5000 + 0.3413 = 0.8413 P(x > 72) = 1 - 0.8413 = 0.1587b.At x = 60P(x 60) = 0.5000 + 0.4986 = 0.9986 P(x < 60) = 1 - 0.9986 = 0.0014 c.At x = 70P(x 70) = 0.5000 + 0.1293 = 0.6293At x = 66P(x 66) = 0.5000 - 0.3413 = 0.1587P(66 x 70) = P(x 70) - P(x 66) = 0.6293 - 0.1587 = 0.4706d.P(x 72) = 1 - P(x > 72) = 1 - 0.1587 = 0.841318.a.Find P(x 60)At x = 60P(x < 60) = 0.5000 + 0.2549 = 0.7549 P(x 60) = 1 - P(x < 60) = 0.2451b.Find P(x 30)At x = 30P(x 30) = 0.5000 - 0.3830 = 0.1170 c.Find z-score so that P(z z-score) = 0.10z-score = 1.28 cuts off 10% in upper tailNow, solve for corresponding value of x.x = 49 + (16)(1.28) = 69.48So, 10% of subscribers spend 69.48 minutes or more reading The Wall Street Journal.19.We have m = 3.5 and s = .8.a.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) = P(z > .63) = 1 - P(z < .63) = 1 - .7357 = .2643The rainfall is less than 3 inches in 26.43% of the Aprils.c.z = 1.28 cuts off approximately .10 in the upper tail of a normal distribution.x = 3.5 + 1.28(.8) = 4.524If it rains 4.524 inches or more, April will be classified as extremely wet.20.We use m = 27 and s = 8a.P(x 11) = P(z -2) = .5000 - .4772 = .0228The probability a randomly selected subscriber spends less than 11 hours on the computer is .025.b.P(x > 40) = P(z > 1.63) = 1 - P(z 1.63) = 1 - .9484 = .05165.16% of subscribers spend over 40 hours 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 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 < 0) = .1772P(400 z 500) = .1772 + .2422 = .4194The probability a worker earns between $400 and $500 is .4194.b.Must find the z-value that cuts off an area of .20 in the upper tail. Using the normal tables, we find z = .84 cuts off approximately .20 in the upper tail.So, x = m + zs = 441.84 + .84(90) = 517.44Weekly earnings of $517.44 or above will put a production worker in the top 20%.c.At 250, P(x 250) = P(z -2.13) = .5000 - .4834 = .0166The probability a randomly selected production worker earns less than $250 per week is .0166.23.a. Area to left is .5000 - .4772 = .0228b.At x = 60 Area to left is .0228At x = 75 Area to left is .3085P(60 x 75) = .3085 - .0228 = .2857c. Area = .5000 - .3413 = .1587Therefore 15.87% of students will not complete on time.(60) (.1587) = 9.522We would expect 9.522 students to be unable to complete the exam in time.24.a. We will use as an estimate of m and s as an estimate of s in parts (b) - (d) below.b.Remember the data are in thousands of shares.At 800P(x 800) = P(z -.90) = 1 - P(z .90) = 1 - .8159 = .1841The probability trading volume will be less than 800 million shares is .1841c. At 1000P(x 1000) = P(z .85) = 1 - P(z .85) = 1 - .8023 = .1977The probability trading volume will exceed 1 billion shares is .1977d.A z-value of 1.645 cuts off an area of .05 in the upper tailx = m + zs = 902.75 + 1.645(114.185) = 1,090.584They should issue a press release any time share volume exceeds 1,091 million.25.a.Find P(x > 100)At x = 100P(x > 100) = P(z .5) = 0.6915b.Find P(x 90)At x = 90P(x 90) = .5000 - .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 = .134127.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 < 10) = 1 - e-10/20 = .3935b.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 ) - (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 < 2) = 1 - e-2/2.78 = .5130b.P(x > 5) = 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 probability 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 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 < 210) = (1 / 25)(210 - 200) = 0.40d.E (x) = (200 + 225)/2 = 212,500If she waits, her expected sale price will be $2,500 higher than if she sells it back to her company now. However, there is a 0.40 probability that she will get less. Its a close call. But, the expected value approach to decision making would suggest she should wait.34.a.For a normal distribution, the mean and the median are equal.m = 63,000b.Find the z-score that cuts off 10% in the lower tail.z-score = -1.28Solving for x, x = 63,000 - 1.28 (15000)= 43,800The lower 10% of mortgage debt is $43,800 or less.c.Find P(x > 80,000)At x = 80,000P(x > 80,000) = 1.0000 - .8708 = 0.1292d.Find the z-score that cuts 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(-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 = -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= Expected 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