Ch06 - Chapter 6 Continuous Probability Distributions...

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6 - 1 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.
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Chapter 6 6 - 2 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
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Continuous Probability Distributions 6 - 3 3. a. 3 / 20 1 / 10 1 / 20 110 120 130 140 f ( x ) x Minutes b. P ( x 130) = (1/20) (130 - 120) = 0.50 c. P ( x > 135) = (1/20) (140 - 135) = 0.25 d. 120 140 ( ) 130 2 Ex + == minutes 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. 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 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
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Chapter 6 6 - 4 e. P ( x 290) = .0386(310.6 - 290) = .7952 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 <+ > == ±²²³ ²² ´± ² ²³ ²²´ Therefore, the probability is .04 + .64 = .68 7. 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. 8. 100 = 10 σ 70 80 90 110 120 130
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Continuous Probability Distributions 6 - 5 9. a. b. .6826 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50. c. .9544 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50. 10. a. .3413 b. .4332 c. .4772 d. .4938 11. a. .3413 These probability values are read directly from the table of areas for the standard b. .4332 normal probability distribution. See Table 1 in Appendix B. c. .4772 d. .4938 e. .4987 12. a. .2967 b. .4418 c. .5000 - .1700 = .3300 d. .0910 + .5000 = .5910 50 = 5 σ 35 40 45 55 60 65 0 -3 -2 -1 +1 +2 +3
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Chapter 6 6 - 6 e. .3849 + .5000 = .8849 f. .5000 - .2611 = .2389 13. a. .6879 - .0239 = .6640 b. .8888 - .6985 = .1903 c. .9599 - .8508 = .1091 14. a. Using the table of areas for the standard normal probability distribution, the area of .4750 corresponds to z = 1.96.
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This note was uploaded on 03/01/2012 for the course ECON 371 taught by Professor Staff during the Spring '08 term at UVA.

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Ch06 - Chapter 6 Continuous Probability Distributions...

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