45-733 Probability and Statistics I (3rd Mini AY 1999-2000 Flex-Mode and Flex-Time)

Assignment #5 Answers


3.26 Let Z = Number of Patients Who Survive. We are given n = 20 and p = .8

  1. P(Z = 14) = F(14) - F(13) = .196 - .087 = .109
  2. P(Z ³ 10) = 1 - F(9) = 1 - .001 = .999
  3. P(14 £ Z £ 18) = F(18) - F(13) = .931 - .087 = .844
  4. P(Z £ 16) = F(16) = .589

3.28 Let Z = Number of Falsified Application Forms. We are given n = 5 and p = .35

P(Z ³ 1) = 1 - F(0) = 1 - .655 = .884
P(Z ³ 2) = 1 - F(1) - F(0) =

        æ5ö
 .884 - ç ÷(.35)1(.65)4
        è1ø

3.30 Let Z = Number of Successful Operations. We are given n = 5.

  1. With p = .8 P(Z = 5) = F(5) - F(4) = 1 - .672 = .328
  2. With p = .6 P(Z = 4) = F(4) - F(3) = .922 - .663 = .259
  3. With p = .3 P(Z < 2) = F(1) = .528

3.82 Let X = Number of Customers Arriving in an Hour. We are given l = 7.

  1. P(X £ 3) = F(3) = .082
  2. P(X ³ 2) = 1 - F(1) = 1 - .007 = .993
  3. P(X = 5) = F(5) - F(4) = .301 - .173 = .128
3.83 Let Y = 10X = Total Service Time in a Given Hour.

E(Y) = 10E(X) = 10*7 = 70
VAR(Y) = 100VAR(X) = 100*7 = 700

3.90 With n = 20 and p = .05, then l = 20*.05 = 1
                                
                          Bin. Poisson
                         ------------
                       0 |.358   .368|  
                         |           |
                       1 |.378   .368|  
                         |           |
                       2 |.189   .184|  
                         |           |
                       3 |.059   .061|  
                         |           |
                       4 |.013   .015|  
                         |           |
                         -------------  
                                        
3.91 Use Poisson Approximation: l = 100*.03 = 3 and let X = The Number of Sales Per 100 Contacts. Hence

P(X ³ 1) = 1 - F(0) = 1 - .05 = .95

4.38 Let X = Bearing Diameter. We are given: X ~ N(3.0005, .0012). The requested probability is:

1 - P(2.998 < X < 3.002) =
1 - P[(2.998 - 3.0005)/.001 < (X - 3.0005)/.001 < (3.002 - 3.0005)/.001) =
1 - P(-2.5 < Z < 1.5) = 1 - [F(1.5) - F(-2.5)] = 1 - [1 - F(-1.5) - F(-2.5)]
=
1 - [1 - .0668 - .0062] = .0668 + .0062 = .0730

4.39 Because the Normal distribution is unimodal and symmetric, setting the mean diameter equal to 3.0000 will maximize the percentage of good ball bearings and minimize the percentage of scrap ball bearings.

4.40 Let X = Grade Point Average. We are given: X ~ N(2.4, .82). The requested probability is:

P(X > 3) = P[(X - 2.4)/.8 > (3 - 2.4)/.8) = P(Z > .75) =
1 - F(.75)
= .2266