ESBE4eAISEsm06 - Chapter 6 Continuous Probability...

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Chapter 6 Continuous Probability Distributions Learning Objectives 1. Understand the difference between how probabilities are computed for discrete and continuous random variables. 2. Know how to compute probability values for a continuous uniform probability distribution and be able to compute the expected value and variance for such a distribution. 3. Be able to compute probabilities using a normal probability distribution. Understand the role of the standard normal distribution in this process. 4. Be able to use the normal distribution to approximate binomial probabilities. 5. Be able to compute probabilities using an exponential probability distribution. 6. Understand the relationship between the Poisson and exponential probability distributions. 6 - 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 6 Solutions: 1. a. 3 2 1 .50 1.0 1.5 2.0 f ( x ) x b. P ( x = 1.25) = 0. The probability of any single point is zero since the area under the curve above any single point is zero. c. P (1.0 x 1.25) = 2(.25) = .50 d. P (1.20 < x < 1.5) = 2(.30) = .60 2. a. .15 .10 .05 10 20 30 40 f ( x ) x 0 b. P ( x < 15) = .10(5) = .50 c. P (12 x 18) = .10(6) = .60 d. 10 20 () 1 5 2 Ex + == e. 2 (20 10) Var( ) 8.33 12 x 3. a. Length of Interval = 310.6 - 284.7 = 25.9 1 for 284.7 310.6 25.9 0 elsewhere x fx ≤≤ = b. Note: 1/25.9 = .0386 6 - 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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Continuous Probability Distributions P(x < 290) = .0386(290 - 284.7) = .2046 c. P ( x 300) = .0386(310.6 - 300) = .4092 d. P (290 x 305) = .0386(305 - 290) = .5790 e. P ( x 290) = .0386(310.6 - 290) = .7952 4. a. 1.5 1.0 .5 1 2 3 f ( x ) x 0 b. P (.25 < x < .75) = 1 (.50) = .50 c. P ( x .30) = 1 (.30) = .30 d. P ( x > .60) = 1 (.40) = .40 5. a. P (10,000 x < 12,000) = 2000 (1 / 5000) = .40 The probability your competitor will bid lower than you, and you get the bid, is .40. b. P (10,000 x < 14,000) = 4000 (1 / 5000) = .80 c. A bid of $15,000 gives a probability of 1 of getting the property. d. Yes, the bid that maximizes expected profit is $13,000. The probability of getting the property with a bid of $13,000 is P (10,000 x < 13,000) = 3000 (1 / 5000) = .60. The probability of not getting the property with a bid of $13,000 is .40. The profit you will make if you get the property with a bid of $13,000 is $3000 = $16,000 - 13,000. So your expected profit with a bid of $13,000 is EP ($13,000) = .6 ($3000) + .4 (0) = $1800. If you bid $15,000 the probability of getting the bid is 1, but the profit if you do get the bid is only $1000 = $16,000 - 15,000. So your expected profit with a bid of $15,000 is EP ($15,000) = 1 ($1000) + 0 (0) = $1,000. 6 - 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 6 6. a. P (12 x 12.05) = .05(8) = .40 b. P ( x 12.02) = .08(8) = .64 c. ( 11.98) ( 12.02) .005(8) .04 .64 .08(8) Px <+ > == 1442 44 31 4 42 443 Therefore, the probability is .04 + .64 = .68 7. a.
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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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ESBE4eAISEsm06 - Chapter 6 Continuous Probability...

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