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Unformatted text preview: ] ‘Practice Questions 4 Answers 1. a. C b .B c. D d. A 2. a. C b. D c. B 3. a. B b. C 4. a. A b. D 5. a. D b. C 6. a. B b. A 7. a. D b. B c. C d. B e. A f. D g. A h. B i. C 8. a. D b. A 9. a. C b. A 10. B 11. B 12. D 13. A 14. C 15. D 16. A 17. C 18. B 19. a. P ( X > 8) = (20-8) / (20-0) = 12 / 20 = 0.60 b. P (10 < X < 15) = (15-10) / (20-0) = 5 / 20 = 0.25 c. The range is a total of 20 minutes. Seventy percent of the time, you should be called in before you wait 14 minutes. 20. a. 0.3859 b. 0.8907 c. 0.6985 d. 0.5549 21. a. k = -0.95 b. k = 0.18 c. k = 1.04 d. k = 1.33 22. P = 0.24, n = 50, μ = E ( X ) = nP = 12, 2 σ = Var( X ) = nP (1-P) = 9.12, σ = 3.02 Without continuity correction P ( X >10) = P ( Z >-0.66) = 0.7454. 23. a. E ( T ) =1/ λ =15.2, then, λ = 0.0658, and P ( T > 45) = 45 2.961 e e λ-- = = 0.0518 b. P ( T < t ) = 0.90, 1- t e λ- = 0.90 ⇒ 0.0658 t e- = 0.10 ⇒-0.0658 t =-2.3026 ⇒ t ; 35 minutes 24. a. 18 to 30 minutes represent 80% of the distance between 15 and 30 minutes. Hence, 18 to 30 minutes represent 80% of the distance between 15 and 30 minutes....
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