ch. 6 solutions

# ch. 6 solutions - 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 compute probabilities using an exponential probability distribution. 5. Understand the relationship between the Poisson and exponential probability distributions. Note to Instructor: 6 - 1

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Chapter 6 For many of the exercises in this chapter, we provide answers obtained using hand calculations and the tables as well as Excel solutions. Small differences due to rounding are often present. 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 ( ) 15 2 E x + = = e. 2 (20 10) Var( ) 8.33 12 x - = = 3. a. 6 - 2
Continuous Probability Distributions 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 E x + = = 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 f x = 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 e. P ( x 290) = .0386(310.6 - 290) = .7952 6 - 3

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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) P x P x < + = = 1 442 4 43 1 442 4 43 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
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## This note was uploaded on 04/22/2008 for the course ISM 2300 taught by Professor Siddiqui during the Spring '08 term at Wayne State University.

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ch. 6 solutions - Chapter 6 Continuous Probability...

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