Eastwood's BA201 Assignments        Draft

Homework Answers for Chapter 5: Exercises 36, 43, 45, 46, 47, 48, 49, 50, 51, 53

36.        m = 13.2, found by 3.00 + 5.20 + 3.50 + 1.50

            s² = 0.86, found by 0.36 + 0.016 + 0.16 + 0.324

40.        The probability of exactly two Spanish-speaking Americans on the jury is 0.168 (Appendix A, n = 12, p = 0.3, and x = 2).  One might argue that this is a sufficiently large chance and agree with the government lawyer. A stronger argument could be made by presenting the probability of two or fewer Spanish-speaking jurors (0.253, found as P(0)+P(1)+P(2) = 0.014+0.071+0.168)

43.        a.        

            b.        

                        P(X ³ 1) = 1 – P(0) = 1 – 0.2917 = 0.7083

45.            

46.        a.         0.60, found by

            b.         0.20, found by

47.        a.         0.0498

            b.         0.7746, found by (1 – 0.0498)⁵

48.        a.         m = 3, probability = 0.0498

            b.         0.5768, found by 1 – (0.0498 + 0.1494 + 0.2240)

49.        m = 4.0 from Appendix C

            a.         0.0183

            b.         0.1954

            c.         0.6289

            d.         0.5665

50.        a.         Referring to Appendix C and finding P(X < 5) = 0.1353 + 0.2707+0.2707+0.1804 + 0.0902 = 0.9473, which is very close to the goal of 0.95.

            b.        

51.        For NASA, m =np  = 25(1/60,000) = 0.0004167

                                     P(X ³ 1) = 1 – 0.9996 = 0.0004167

            For Air Force, m =25(1/35) = 0.7143

                          P(X ³ 1) = 1 – 0.4895 = 0.5105

            Summarizing, Air Force estimate is 0.5105, and NASA estimate is 0.0004.

53.        Let m =np =155(1/3709) = 0.04179

                     Very Unlikely!

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