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ch. 5 solutions - Chapter 5 Discrete Probability...

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Chapter 5 Discrete Probability Distributions Learning Objectives 1. 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, and understand how an Excel worksheet can be used to ease the burden of calculations. 4. Be able to compute probabilities using a binomial probability distribution and be able to compute these probabilities using Excel’s BINOMDIST function. 5. Be able to compute probabilities using a Poisson probability distribution and be able to compute these probabilities using Excel’s POISSON function. 6. Know when and how to use the hypergeometric probability distribution and be able to compute probabilities using Excel’s HYPGEOMDIST function. Solutions: 5 - 1
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Chapter 5 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 tosses c. Outcome Values of x (H,H) 2 (H,T) 1 (T,H) 1 (T,T) 0 d. 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. Continuous 3. Let Y = position is offered N = position is not offered a. 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 N 3 2 2 1 2 1 1 0 4. 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 Required 2 3 4 3 4 5 6. a. values: 0,1,2,...,20 discrete b. values: 0,1,2,... discrete c. values: 0,1,2,...,50 discrete d. values: 0 x 8 continuous e. values: x > 0 continuous 5 - 2
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Discrete Probability Distributions 7. a. f ( x ) 0 for all values of x . Σ f ( x ) = 1 Therefore, it is a proper probability distribution. b. Probability x = 30 is f (30) = .25 c. Probability x 25 is f (20) + f (25) = .20 + .15 = .35 d. Probability x > 30 is f (35) = .40 8. a. x f ( x ) 1 3/20 = .15 2 5/20 = .25 3 8/20 = .40 4 4/20 = .20 Total 1.00 b. .1 .2 .3 .4 f ( x ) x 1 2 3 4 c. f ( x ) 0 for x = 1,2,3,4. Σ f ( x ) = 1 9. a. Age Number of Children f ( x ) 6 37,369 0.018 7 87,436 0.043 8 160,840 0.080 9 239,719 0.119 10 286,719 0.142 5 - 3
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Chapter 5 11 306,533 0.152 12 310,787 0.154 13 302,604 0.150 14 289,168 0.143 2,021,175 1.001 b. c. f ( x ) 0 for every x Σ f ( x ) = 1 Note: Σ f ( x ) = 1.001 in part (a); difference from 1 is due to rounding values of f ( x ). 10. a. x f ( x ) 1 0.05 2 0.09 3 0.03 4 0.42 5 0.41 1.00 b. x f ( x ) 1 0.04 2 0.10 3 0.12 4 0.46 5 0.28 1.00 5 - 4 6 7 8 9 10 11 12 13 14 .02 .04 .06 .08 .10 .12 .14 .16 f ( x ) x
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Discrete Probability Distributions c. P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83 d. Probability of very satisfied: 0.28 e. 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 Call x f ( x ) 1 0.25 2 0.25 3 0.25 4 0.25 1.00 b. 0.10 0.20 0.30 f ( x ) x 1 2 3 4 0 c. f ( x ) 0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00 d. f (3) = 0.25 e. P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50 12. a. Yes; f ( x ) 0 for all x and Σ f ( x ) = .15 + .20 + .30 + .25 + .10 = 1 b. 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 Σ f ( x ) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1 b. f (2) = 2/6 = .333 c. f (2) + f (3) = 2/6 + 3/6 = .833 14. a. f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150) = 1 - .95 = .05 This is the probability MRA will have a $200,000 profit.
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  • Spring '08
  • Siddiqui
  • Probability theory, Binomial distribution, Discrete probability distribution, Discrete Probability Distributions

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