ESBE4eAISEsm05 - 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. 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. 5 - 1 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Chapter 5 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 5 - 2 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Discrete Probability Distributions d. values: 0 x 8 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. 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 10. a. 5 - 3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
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Chapter 5 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 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.
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This note was uploaded on 05/26/2010 for the course ACC 251 taught by Professor Carl during the Winter '09 term at University of Central Arkansas.

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ESBE4eAISEsm05 - Chapter 5 Discrete Probability...

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