Ch05 - Chapter 5 Discrete Probability Distributions...

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5 - 1 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. 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.
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Chapter 5 5 - 2 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 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
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Discrete Probability Distributions 5 - 3 continuous e. values: x > 0 continuous 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 1 234 c. f ( x ) 0 for x = 1,2,3,4. Σ f ( x ) = 1 9. a. Age Number of Children f ( x )
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Chapter 5 5 - 4 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 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 ) 6 7 8 91 01 11 21 3 14 .02 .04 .06 .08 .10 .12 .14 .16 f ( x ) x
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Discrete Probability Distributions 5 - 5 1 0.04 2 0.10 3 0.12 4 0.46 5 0.28 1.00 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 1 234 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
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Chapter 5 5 - 6 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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Ch05 - Chapter 5 Discrete Probability Distributions...

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