MATH 511, Meade                                                                                             HW Solutions

3.2 –3 (omit c), 5, 14, 18, **3c, **8                                                                           2/13/04

 

3.

    a.  f(x) = x/10, x = 1,2,3,4

         E(x) = (1) (1/10) + (2) (2/10) + (3) (3/10) + (4) (4/10)

                 = (1 + 4  + 9 + 16 )/10 = 30/10 = 3

     b. f(x) = x/55 , x = 1, 2, 3, 4 ,5 , 6, 7, 8, 9, 10

         E(x) =  x=1∑10 x2/55 = [(10)(11)(21)/(6)]/55 = 385/55 = 7

  **c. f(x) = 3(1/4)x ,  x = 1, 2, 3, …

         E(x) = x=1∑∞ 3x (1/4)x = x=1∑∞ 3x/4x  ≤ x=1∑∞ 3x/4x = 3

         This assures that the series does indeed converge. The series,  x=1∑∞ x/4x ,

         can now be approximated. After about ten approximations it is clear that

         the series converges to 0.44444… which is 4/9.

         So the E(X) = 3(4/9) = (4/3)

     d.  f(x) = (1/30)(x+1)2 ,  x = 0, 1, 2, 3

          E(x) = (1/30) [ 0 + 4 + 18 + 48] = 70/30 = 7/3

     e. f(x) = 2/n(n+1)x ,   x = 1, 2, 3, …, n

         E(x) =2/n(n+1) i=1∑n x2 = [2/n(n+1)] [n(n+1)(2n+1)/6)] = (2n+1)/3

 

5. E(x) = [2,987,994(0) + 12,000(25) + 4(10,000) + 1(50,000) + 1(200,000)] - 0.50

                                                   3,000,000

            = 19 ⅔ cents – 50 cents = - 30⅓ cents

 

**8. f(x) = 6/( p2x2),  x = 1, 2, 3, …

    E(x) = (6/ p2) x=1∑∞ x/x2 = (6/ p2) x=1∑∞ 1/x ; this is the harmonic series, which

               diverges. Therefore E(x) does not exist. One way to show this is the

               integral test for infinite series.

 

14.

      a. [16(25) + 3(100) + 1(300)]/20 = 1,000/20 = 50. Note that this makes sense: 1000 students divided into 20 classes; the average class size must be 1000/20=50.

      b. f(x) = 0.4, x = 25

                = 0.3, x = 100

                = 0.3, x = 300.  Note that here you are selecting a random student, not a random class.

      c. E(x) = 25(0.4) + 100(0.3) + 300(0.3) = 130

 

18.

      a. E[(X - m) / s] = (1/s) E(X - m) = (1/s) [E(X) – E(m)] =(1/s) [m – m] = 0

      b. E{[(X – m) / s]2} = (1/s2) (E[(X – m)2]) = (1/s2) (E(X2) – 2mE(X) + m2)

                                   = (1/s2) (E(X2) – 2m2 + m2) = (1/s2) (E(X2) – m2)

                                   = (1/s2) s2  = 1

 

 

 

 

3.3 – 3, 5, 7(abc), 14, **7(de)

 

3. There is six-question multiple choice test with five possible answers, only one

     of which is correct. If a student guesses randomly and independently, then

     what is the probability that:

a.      The student gets questions 1 and 4 correct.

(0.2)2(0.8)4 = 0.0164

    b.  The student gets exactly two questions right. 6C2 (0.2)2(0.8)4 = 0.2458

 

5. The number of people who believe that the IRS abuses their power has a

    binomial distribution. b(25,0.7). In order to solve this problem using Appendix

    C, Table II, one needs to convert the problems so that p ≤ 0.5.

    So Let Y = the number who do not believe the IRS abuses their power. Y has

    a distribution of b(25, 0.3).

   

    a. P(X ≥ 13) = P(Y ≤ 12) = 0.9825

    b. P(X ≤ 11) = P(Y ≥ 14) = 1 – P(Y < 14) = 1 – P(Y ≤ 13) = 1 – 0.9940 = 0.0060

    c. P(X = 12) = 25C12 (0.7)12(0.3)13 = 0.0115

    d. m(x) = np = (25)(0.7) = 17.5

        s2 (x) = np(1 – p) = (25)(0.7)(0.3) = 5.25

        s(x) = √s2 = 2.29

 

7.

    a. W has a binomial distribution b(2000, p/4). The probabilty comes from that

        the fact that exactly one fourth of the unit circle sits inside the unit square.

        The area of the unit circle is Acircle(r = 1) = p(1)2 = p.

    b. m(w) = np = (2000)( p/4) = 1,570.796

        s2 (w) = np(1 – p) = (2000)( p/4)(1 - p/4)= 337.096

        s(w)= √s2 = 18.360

    c. E(W/500) =  m(w)/500 = p

 **d. Click on the Excel link on the solutions column for a spreadsheet concerning

        parts d and e of this problem.

 **e. Notice that exactly one eighth of the unit sphere is contained in the unit cube

        The volume of the unit sphere is (4/3) p. So the probability that a point on

        The unit cube is also on the unit sphere is (1/8) (4/3) p = p/6

 

14.

      a. X = b(8, 0.90), Let Y = number of mints weighing less than 20.7g = b(8, 0.10)

      b.  

          i. P(X = 8) = 8C8 (0.90)8(0.10)0 = 0.4305

          ii. P(X ≤ 6) = P(Y ≥ 2) = 1 – P(Y < 2) = 1 – P(Y ≤ 1) = 1 – 0.8131 = 0.1869

          iii. P(X ≥ 6) = P(Y ≤ 2 = 0.9619