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= 2.0