商务与经济统计习题答案（第8版中文版）SBE8-SM05.doc

Discrete Probability DistributionsChapter 5Discrete Probability DistributionsLearning Objectives1.Understand the concepts of a random variable and a probability distribution.2.Be able to distinguish between discrete and continuous random variables.3.Be able to compute and interpret the expected value, variance, and standard deviation for a discrete random variable.4.Be able to compute and work with probabilities involving a binomial probability distribution.5.Be able to compute and work with probabilities involving a Poisson probability distribution.6.Know when and how to use the hypergeometric probability distribution.Solutions:1.a.Head, Head (H,H)Head, Tail (H,T)Tail, Head (T,H)Tail, Tail (T,T)b.x = number of heads on two coin tossesc.OutcomeValues of x(H,H)2(H,T)1(T,H)1(T,T)0d.Discrete. It may assume 3 values: 0, 1, and 2.2.a.Let x = time (in minutes) to assemble the product.b.It may assume any positive value: x > 0.c.Continuous3.LetY = position is offeredN = position is not offereda.S = (Y,Y,Y), (Y,Y,N), (Y,N,Y), (Y,N,N), (N,Y,Y), (N,Y,N), (N,N,Y), (N,N,N)b.Let N = number of offers made; N is a discrete random variable.c.Experimental Outcome (Y,Y,Y) (Y,Y,N) (Y,N,Y) (Y,N,N) (N,Y,Y) (N,Y,N) (N,N,Y) (N,N,N)Value of N322121104.x = 0, 1, 2, . . ., 12.5.a.S = (1,1), (1,2), (1,3), (2,1), (2,2), (2,3)b.Experimental Outcome(1,1)(1,2)(1,3)(2,1)(2,2)(2,3)Number of Steps Required2343456.a.values:0,1,2,.,20discreteb.values:0,1,2,.discretec.values:0,1,2,.,50discreted.values:0 x 8continuouse.values:x > 0continuous7.a.f (x) 0 for all values of x.S f (x) = 1 Therefore, it is a proper probability distribution.b.Probability x = 30 is f (30) = .25c.Probability x 25 is f (20) + f (25) = .20 + .15 = .35d.Probability x > 30 is f (35) = .408.a.xf (x)13/20 = .1525/20 = .2538/20 = .4044/20 = .20Total 1.00b.c.f (x) 0 for x = 1,2,3,4.S f (x) = 19.a.xf (x)115/462= 0.032232/462= 0.069384/462= 0.1824300/462= 0.650531/462= 0.067b.c.All f (x) 0S f (x) = 0.032 + 0.069 + 0.182 + 0.650 + 0.067 = 1.00010.a.xf(x)10.0520.0930.0340.4250.411.00b.xf(x)10.0420.1030.1240.4650.281.00c.P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83d.Probability of very satisfied: 0.28e.Senior executives appear to be more satisfied than middle managers. 83% of senior executives have a score of 4 or 5 with 41% reporting a 5. Only 28% of middle managers report being very satisfied.11.a.Duration of Callxf(x)10.2520.2530.2540.251.00b.c.f (x) 0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00d.f (3) = 0.25e.P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50 12.a.Yes; f (x) 0 for all x and S f (x) = .15 + .20 + .30 + .25 + .10 = 1b.P(1200 or less)= f (1000) + f (1100) + f (1200)= .15 + .20 + .30= .65 13.a.Yes, since f (x) 0 for x = 1,2,3 and S f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1b.f (2) = 2/6 = .333c.f (2) + f (3) = 2/6 + 3/6 = .83314.a.f (200)= 1 - f (-100) - f (0) - f (50) - f (100) - f (150)= 1 - .95 = .05This is the probability MRA will have a $200,000 profit.b.P(Profit)= f (50) + f (100) + f (150) + f (200)= .30 + .25 + .10 + .05 = .70c.P(at least 100)= f (100) + f (150) + f (200)= .25 + .10 +.05 = .4015.a.xf (x)x f (x)3.25.756 .503.009 .25 2.251.006.00E (x) = m = 6.00b.xx - m(x - m)2f (x)(x - m)2 f (x)3-39.252.256 00.500.009 39.252.254.50Var (x) = s2 = 4.50c.s = = 2.1216.a.yf (y)y f (y)2.20.404 .301.207 .402.808 .10 .801.005.20E(y) = m = 5.20b.yy - m(y - m)2f (y)(y - m)2 f (y)2-3.2010.24.202.0484 -1.20 1.44.30 .4327 1.80 3.24.401.2968 2.80 7.84.10 .7844.56017.a/b.xf (x)x f (x)x - m(x - m)2(x - m)2 f (x)0.10.00-2.456.0025.6002501.15 .15-1.452.1025 .3153752.30 .60- .45 .2025 .0607503.20 .60 .55 .3025 .0605004.15 .60 1.552.4025 .3603755.10 .50 2.556.5025 .6502502.452.047500E (x)= m = 2.45s2= 2.0475s= 1.430918.a/b.xf (x)x f (x)x - m(x - m)2(x - m)2 f (x)0.01 0-2.35.29 0.05291.23 .23-1.31.69 0.38872.41 .82-0.30.09 0.03693.20 .60 0.70.49 0.0984.10 .40 1.72.89 0.2895.05 .25 2.77.29 0.36452.31.23E(x) = 2.3 Var(x) = 1.23The expected value, E (x) = 2.3, of the probability distribution is the same as that reported in the 1997 Statistical Abstract of the United States. 19.a.E (x) = S x f (x) = 0 (.50) + 2 (.50) = 1.00b.E (x) = S x f (x) = 0 (.61) + 3 (.39) = 1.17c.The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will make more points in the long run with the 3 - point shot.20.a.xf (x)x f (x)0.90 0.00400 .04 16.001000 .03 30.002000 .01 20.004000 .01 40.006000 .01 60.001.00166.00E (x) = 166. If the company charged a premium of $166.00 they would break even.b.Gain to Policy Holderf (Gain)(Gain) f (Gain)-260.00.90-234.00140.00.04 5.60740.00.03 22.201,740.00.01 17.403,740.00.01 37.405,740.00.01 57.40 -94.00E (gain) = -94.00. The policy holder is more concerned that the big accident will break him than with the expected annual loss of $94.00.21.a.E (x)= S x f (x)= 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5)=4.05b.E (x)= S x f (x)= 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5)=3.84c.Executives:s2 =S (x - m)2 f(x) = 1.2475Middle Managers: s2 =S (x - m)2 f(x) = 1.1344d.Executives: s = 1.1169Middle Managers:s = 1.0651e.The senior executives have a higher average score: 4.05 vs. 3.84 for the middle managers. The executives also have a slightly higher standard deviation.22.a.E (x)= S x f (x)= 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445The monthly order quantity should be 445 units.b.Cost:445 $50=$22,250Revenue:300 $70= 21,000$ 1,250 Loss23.a.Laptop: E (x) = .47(0) + .45(1) + .06(2) + .02(3) = .63Desktop: E (x) = .06(0) + .56(1) + .28(2) + .10(3) = 1.42b.Laptop: Var (x) = .47(-.63)2 + .45(.37)2 + .06(1.37)2 + .02(2.37)2 = .4731Desktop: Var (x) = .06(-1.42)2 + .56(-.42)2 + .28(.58)2 + .10(1.58)2 = .5636c.From the expected values in part (a), it is clear that the typical subscriber has more desktop computers than laptops. There is not much difference in the variances for the two types of computers.24.a.Medium E (x)= S x f (x)= 50 (.20) + 150 (.50) + 200 (.30) = 145Large: E (x)= S x f (x)= 0 (.20) + 100 (.50) + 300 (.30) = 140Medium preferred.b.Mediumxf (x)x - m(x - m)2(x - m)2 f (x) 50.20-9590251805.0150.50 5 25 12.5200.30 553025 907.5s2 =2725.0Largeyf (y)y - m(y - m)2(y - m)2 f (y) 0.20-140196003920100.50 -40 1600 800300.30 160256007680 s2 = 12,400Medium preferred due to less variance.25.a.b.c.d.e.P (x 1) = f (1) + f (2) = .48 + .16 = .64f.E (x) = n p = 2 (.4) = .8Var (x) = n p (1 - p) = 2 (.4) (.6) = .48s = = .692826.a.f (0) = .3487b.f (2) = .1937c.P(x 2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298d.P(x 1) = 1 - f (0) = 1 - .3487 = .6513e.E (x) = n p = 10 (.1) = 1f.Var (x) = n p (1 - p) = 10 (.1) (.9) = .9s = = .948727.a.f (12) = .1144b.f (16) = .1304 c.P (x 16)= f (16) + f (17) + f (18) + f (19) + f (20)= .1304 + .0716 + .0278 + .0068 + .0008= .2374d.P (x 15) = 1 - P (x 16) = 1 - .2374 = .7626e.E (x) = n p = 20(.7) = 14f.Var (x) = n p (1 - p) = 20 (.7) (.3) = 4.2s = = 2.049428.a.b.P(at least 2)=1 - f(0) - f(1)=1 - .0905 - .2673 = .6422c.29.P(At Least 5)= 1 - f (0) - f (1) - f (2) - f (3) - f (4)= 1 - .0000 - .0005 - .0031 - .0123 - .0350= .949130.a.Probability of a defective part being produced must be .03 for each trial; trials must be independent.b.Let:D = defectiveG = not defectivec.2 outcomes result in exactly one defect.d.P (no defects) = (.97) (.97) = .9409P (1 defect) = 2 (.03) (.97) = .0582P (2 defects) = (.03) (.03) = .000931.Binomial n = 10 and p = .05a.Yes. Since they are selected randomly, p is the same from trial to trial and the trials are independent.b.f (2) = .0746c.f (0) = .5987d.P (At least 1) = 1 - f (0) = 1 - .5987 = .401332.a.90b.P (at least 1) = f (1) + f (2)Alternativelyc.P (at least 1) = 1 - f (0)d.Yes; P (at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack would be catastrophic.33.a.Using the binomial formula or the table of binomial probabilities with p = .5 and n = 20, we find:xf(x)120.1201130.0739140.0370150.0148160.0046170.0011180.0002190.0000200.00000.2517The probability 12 or more will send representatives is 0.2517.b.Using the binomial formula or the tables, we find:xf(x)00.000010.000020.000230.001140.004650.01480.0207c.E(x) = n p = 20(0.5) = 10d.s2 = n p (1 - p) = 20(0.5)(0.5) = 5s = = 2.236134.a.f (3) = .0634 (from tables)b.The answer here is the same as part (a). The probability of 12 failures with p = .60 is the same as the probability of 3 successes with p = .40.c.f (3) + f (4) + + f (15)=1 - f (0) - f (1) - f (2)=1 - .0005 - .0047 - .0219= .972935.a.f (0) + f (1) + f (2) = .0115 + .0576 + .1369 = .2060b.f (4) = .2182c.1 - f (0) + f (1) + f (2) + f (3) = 1 - .2060 - .2054 = .5886d.m = n p = 20 (.20) = 436.xf (x)x - m(x - m)2(x - m)2 f (x)0.343-.9.81.277831.441 .1 .01.004412.1891.11.21.228693.0272.14.41.11907 1.000s2 = .6300037.E(x) = n p = 30(0.29) = 8.7s2 = n p (1 - p) = 30(0.29)(0.71) = 6.177s = = 2.48538.a.b.c. d.P (x 2) = 1 - f (0) - f (1) = 1 - .0498 - .1494 = .800839.a.b.m = 6 for 3 time periodsc.d.e.f.40.a.m = 48 (5/60) = 4b.m = 48 (15 / 60) = 12c.m = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes.The probability none will be waiting after 5 minutes is .0183.d.m = 48 (3 / 60) = 2.4The probability of no interruptions in 3 minutes is .0907.41.a.30 per hourb.m = 1 (5/2) = 5/2c.42.a.b.For a 3-month period: m = 1c.For a 6-month period: m = 2The probability of 1 or more flights = 1 - f (0) = 1 - 0.1353 = 0.864743.a.b.f (0) + f (1) + f (2) + f (3)f (0) = .000045 (part a)Similarly, f (2) = .00225, f (3) = .0075and f (0) + f (1) + f (2) + f (3) = .010245c.2.5 arrivals / 15 sec. period Use m = 2.5d.1 - f (0) = 1 - .0821 = .917944.Poisson distribution appliesa.m = 1.25 per monthb.c.d.P (More than 1) = 1 - f (0) - f (1) = 1 - 0.2865 - 0.3581 = 0.355445.a.For 1 week, m = 450 / 52 = 8.65b.c.For a 1-day period: m = 450 / 365 = 1.23Probability of 2 or more deaths = 1 - f (0) - f (1) = 1 - 0.2923 - 0.3595 = 0.348246.a.b.c.d.47.48.Hypergeometric with N = 10 and r = 6a.b.Must be 0 or 1 prefer Coke Classic.P (Majority Pepsi) = f (1) + f (0) = .333349.Parts a, b & c involve the hypergeometric distribution with N = 52 and n = 2a.r = 20, x = 2b.r = 4, x = 2c.r = 16, x = 2d.Part (a) provides the probability of blackjack plus the probability of 2 aces plus the probability of two 10s. To find the probability of blackjack we subtract the probabilities in (b) and (c) from the probability in (a).P (blackjack) = .1433 - .0045 - .0905 = .048350.N = 60 n = 10a.r = 20 x = 0f (0)= = .01b.r = 20 x = 1f (0)= .07c.1 - f (0) - f (1) = 1 - .08 = .92d.Same as the probability one will be from Hawaii. In part b that was found to equal approximately .07.51.a.b.c.d.52.Hypergeometric with N = 10 and r = 2.Focus on the probability of 0 defectives, then the probability of rejecting the shipment is 1 - f (0).a.n = 3, x = 0P (Reject) = 1 - .4667 = .5333b.n = 4, x = 0P (Reject) = 1 - .3333 = .6667c.n = 5, x = 0P (Reject) = 1 - .2222 = .7778d.Continue the process. n = 7 would be required with the probability of rejecting = .933353.a., b. and c.xf (x)x f (x)x - m(x - m)2(x - m)2 f (x)10.180.18-2.305.290.952220.180.36-1.301.690.608430.030.09-0.300.090.008140.381.520.700.490.744850.231.151.702.893.32351.003.305.6370E(x) = m = 3.30 s2 = 5.6370s = = 2.374254.a. and b.xf (