course1solutions_0503

# course1solutions_0503 - Course 1 May 2003 Answer Key 1. 2....

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Course 1 p. 1 of 19 May 2003 Course 1 May 2003 Answer Key 1. D 21. D 2. A 22. B 3. E 23. A 4. D 24. C 5. B 25. B 6. A 26. A 7. C 27. D 8. B 28. C 9. D 29. E 10. E 30. E 11. B 31. D 12. D 32. C 13. C 33. B 14. C 34. C 15. C 35. E 16. B 36. D 17. C 37. D 18. D 38. C 19. A 39. E 20. C 40. B

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Course 1 p. 2 of 19 May 2003 1. Solution: D Let event that a viewer watched gymnastics event that a viewer watched baseball event that a viewer watched soccer G B S = = = Then we want to find () () () () ( ) ( ) ( ) ( ) Pr 1 Pr 1P r P r P r P r P r P r P r 1 0.28 0.29 0.19 0.14 0.10 0.12 0.08 1 0.48 0.52 c GBS GBSG BG SB SG B S  ∪∪ =−  + + ∩ − ∩ + ∩ ∩ = ++−−−+ = −= 2. Solution: A The graph in A contains the graphs of the functions ( ) 3 f xx = and 6 f ′′ = . More generally (without making any assumptions regarding the exact definition of the functions f x ), one can reason as follows: (E) is out because the second derivative of a linear function is identically 0. (B) and (D) are out because the curve which is non-linear and would have to be ( ) f x is increasing at an increasing rate in the first quadrant. This says ( ) f x is positive and increasing which means f x must be positive for 0 x > . (C) is out because the curve which would have to be ( ) f x is decreasing at a decreasing rate in the second quadrant. Therefore ( ) f x would have to be negative and increasing which implies f x must be positive when 0 x < . 3. Solution: E Observe () () ( ) ( ) 22 00 0 lim lim lim lim lim x cf x dg x cf x dg x c d c d cd fx gx fx gx →→ −− + == = = + (Note L’Hôspital’s Rule does not apply in this problem because the limit in the denominator is not 0.)
Course 1 p. 3 of 19 May 2003 4. Solution: D The distribution function of an exponential random variable T with parameter θ is given by ( ) 1, 0 t Ft e t = −> Since we are told that T has a median of four hours, we may determine as follows: () 4 4 1 41 2 1 2 4 ln 2 4 ln 2 Fe e == = −= = Therefore, ( ) ( ) 5ln 2 55 4 4 Pr 5 1 5 2 0.42 TF e e −− ≥= = = = = 5. Solution: B Let event that customer insures more than one car event that customer insures a sports car M S = = Then applying DeMorgan’s Law, we may compute the desired probability as follows: ( ) ( ) () ( ) ()() ( ) ( ) Pr Pr 1 Pr 1 Pr Pr Pr 1 Pr Pr Pr Pr 1 0.70 0.20 0.15 0.70 0.205 c cc M SM S M S M S M S MSS M M  ∩= = = +  =− + + =

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Course 1 p. 4 of 19 May 2003 6. Solution: A The calculation requires integrating over the indicated region. () 21 12 1 1 1 22 2 2 2 2 4 5 00 0 0 0 2 1 1 1 1 23 3 3 4 5 000 0 0 2 1 2 3 3 5 84 4 4 4 44 33 3 5 5 88 8 5 6 5 6 5 6 8 3 9 9 9 45 45 888 5 6 8 399 9 x x x x x x x x x x x x E X x y dy dx x y dx x x x dx x dx x E Y xy dy dx xy dy dx x x x dx x dx x E X Y x y d y d xx y d x x d x x d == = = = = === = = = = = ∫∫ ( ) ( ) ( ) 11 56 28 54 27 28 56 4 Cov , 0.04 27 45 5 x XY E XY E X EY  =− = =   7. Solution: C Let 2 ux = . Then () ( ) () () 24 2 4 0 2 2 f xd x f u d u f ud u u + 1 35 4 2 = +=
Course 1 p. 5 of 19 May 2003 8. Solution: B Apply Bayes’ Formula. Let Event of an accident A = 1 B = Event the driver’s age is in the range 16-20 2 B = Event the driver’s age is in the range 21-30 3 B = Event the driver’s age is in the range 30-65 4 B =

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## This note was uploaded on 03/27/2012 for the course MATHEMATIC 4523 taught by Professor Zhangqiang during the Spring '12 term at City University of Hong Kong.

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course1solutions_0503 - Course 1 May 2003 Answer Key 1. 2....

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