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statistics for business and economics 11th edition chapter 6 - exercise solution

Statistics for business and economics 11th edition chapter 6 - exercise solution

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Chapter 6 Continuous Probability Distributions 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 - = = 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 Rounding up, we conclude that 80 of the top 100 golfers drive the ball this far. 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 17. Let x = debt amount μ = 15,015, σ = 3540 a. 18,000 15,015 .84 3540 z - = =
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Chapter 6 P ( x > 18,000) = 1- P ( z .84) = 1 - .7995 = .2005 b. 10,000 15,015 1.42 3540 z - = = - P ( x < 10,000) = P ( z < -1.42) = .0778 c. At 18,000, z = .84 from part (a) At 12,000, 12,000 15,015 .85 3540 z - = = - P (12,000 < x < 18,000) = P (-.85 < z < .84) = .7995 - .1977 = .6018 d. 14,000 15,015 .29 3540 z - = = - P ( x 14,000) = P ( z -.29) = .3859 19. We have μ = 3.5 and σ = .8. a. 5.0 3.5 1.88 .8 z - = P ( x > 5.0) = P ( z > 1.88) = 1 - P ( z 1.88) = 1 - .9699 = .0301 The rainfall exceeds 5 inches in 3.01% of the Aprils. b. 3 3.5 .63 .8 z - = ≈ - P ( x < 3.0) = P ( z < -.63) = .2643 The rainfall is less than 3 inches in 26.43% of the Aprils. c. z = 1.28 cuts off approximately .10 in the upper tail of a normal distribution. x = 3.5 + 1.28(.8) = 4.524 If it rains 4.524 inches or more, April will be classified as extremely wet. 20. μ = 77 and σ = 20 a. At x = 50, 50 77 1.35 20 z - = = - P ( z < -1.35) = .0885 So, P ( x < 50) = .0885 b. At x = 100, 100 77 1.15 20 z - = = P ( z > 1.15) = 1 - .8749 = .1251 So, P ( x > 100) = .1251 12.51% of workers logged on over 100 hours. c. A z -value of .84 cuts off an area of approximately .20 in the upper tail. x = μ + z σ = 77 + 20(.84) = 93.8 A worker must spend 93.8 or more hours logged on to be classified a heavy user.
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