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Unformatted text preview: 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. Solutions: 6  1 Chapter 6 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 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 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 Chapter 6 Rounding up, we conclude that 80 of the top 100 golfers drive the ball this far. 6. a. P ( x ≥ 25) = 1 8 (26 – 25) = .125 b. P (21 ≤ x ≤ 25) = 1 8 (25 – 21) = .50 c. This occurs when programming is 20 minutes or less P ( x ≤ 20) = 1 8 (20 – 18) = .25 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....
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This note was uploaded on 11/08/2011 for the course MAT/FIN 272 taught by Professor Burns during the Spring '11 term at Central Connecticut State University.
 Spring '11
 Burns

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