c1solutions_0501 - Course 1 May 2001 Answer Key 1 2 3 4 5 6...

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Course 1 Solutions 1 May 2001 Course 1 May 2001 Answer Key 1 E 21 E 2 C 22 C 3 C 23 A 4 E 24 E 5 C 25 B 6 D 26 B 7 B 27 A 8 A 28 C 9 C 29 D 10 E 30 B 11 D 31 C 12 D 32 E 13 E 33 D 14 A 34 A 15 D 35 C 16 A 36 B 17 B 37 E 18 D 38 D 19 B 39 A 20 D 40 B A7 B8 C8 D9 E8
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Course 1 Solutions 2 May 2001 1. E We are given that 56 bb ep p e −− === It follows that 6 65 5 5 6 5 ln 6 ln ln b b b e ee e e e b b + == = =  =−   = 2. C First, solve for m such that () ( ) 1 1 1.07 1.07 1 500 8 8 1.07 ... 8 1.07 8 8 1 1.07 0.07 5.375 1.07 ln 5.375 ln 1.07 ln 5.375 24.86 ln 1.07 mm m m m m  =+ + + = =   = = We conclude that 25 m = . 3. C Observe that 2 cos / 2 and 2 cost 2 sin dx dy tt t dt dt Therefore, /2 2 2 cos /4 2 2 2 cos / 2 sin / 2 t t dx dt dy dt π πππ = = = =−= It follows that the length of the velocity vector at time 2 t = is given by 2 2 2 22 ππ +− = + .
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Course 1 Solutions 3 May 2001 4. E Let X 1 , X 2 , X 3 , and X 4 denote the four independent bids with common distribution function F . Then if we define Y = max ( X 1 , X 2 , X 3 , X 4 ), the distribution function G of Y is given by () [] 1234 4 4 Pr Pr Pr Pr Pr Pr 13 5 1 sin , 16 2 2 Gy Y y Xy Fy yy π =≤ = ∩≤   = =+ It then follows that the density function g of Y is given by () () ( ) 3 3 ' 1 1 sin cos 4 35 cos 1 sin , 42 2 gy G y y ππ = Finally, [] () 5/2 3/2 3 cos 1 sin 4 EY yg ydy y d y =
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Course 1 Solutions 4 May 2001 5. C The domain of X and Y is pictured below. The shaded region is the portion of the domain over which X <0.2 . Now observe [] () () ( ) 1 0.2 1 0.2 2 00 0 0 0.2 0.2 22 2 0.2 23 3 0.2 0 0 1 Pr 0.2 6 1 6 2 11 6 1 1 1 6 1 1 1 6 1 1 0.8 1 2 x x Xx y d y d x y x y y d x x x x x dx x x dx xd x x  =− + =   =  = = + ∫∫ < | 0.488 = 6. D Let S = Event of a standard policy F = Event of a preferred policy U = Event of an ultra-preferred policy D = Event that a policyholder dies Then [] [ ] [ ] [ ] [ ] [ ] [ ] ( ) ( ) ( ) | | || | 0.001 0.10 0.01 0.50 0.005 0.40 0.001 0.10 0.0141 PDUPU PU D PDSPS PDFPF PDUPU = ++ = =
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Course 1 Solutions 5 May 2001 7. B Let us first determine k : 11 1 1 21 0 00 0 0 1 1 22 2 2 kk kxdxdy kx dy dy k == = = = ∫∫ | Then [] [ ] [ ] [ ] 1 3 1 0 0 1 0 0 1 1 23 1 0 0 0 0 33 2 2 1 66 3 12 1 Cov , 0 3 E X x dydx x dx x E Y y x dxdy ydy y EX Y xy d x d y xy d y y d y y XY EXY EX EY = = = = = =  = =− =−=   | | | | (Alternative Solution) Define g ( x ) = kx and h ( y ) = 1 . Then f ( x , y ) = g ( x ) h ( x ) In other words, f ( x , y ) can be written as the product of a function of x alone and a function of y alone. It follows that X and Y are independent. Therefore, Cov[ X , Y ] = 0 . 8. A By the chain rule, 1 1 100 50 50 dp d dx dy xy x y x y dt dt dt dt −−  +  At the time t 0 in question, we are told that 1 2 , 1 , 3 , and 2 dx dy dt dt = = Therefore, () 0 32 1 50 1 50 40.8 2 tt dp dt = =+ =
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Course 1 Solutions 6 May 2001 9. C The Venn diagram below summarizes the unconditional probabilities described in the problem. In addition, we are told that [] 1 | 3 0.12 PA B C x PA B CA B PA B x ∩∩ =∩ = = ∩+ It follows that () 11 0.12 0.04 33 2 0.04 3 0.06 xx x x x =+ = + = = Now we want to find | 1 1 1 3 0.10 3 0.12 0.06 1 0.10 2 0.12 0.06 0.28 0.467 0.60 c c c c PABC PABC A PA  ∪∪  = −∪ = −−− = −− ==
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Course 1 Solutions 7 May 2001 10. E Let W
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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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c1solutions_0501 - Course 1 May 2001 Answer Key 1 2 3 4 5 6...

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