Course1solutions_1101

# Course1solutions_1101 - Course 1 Solutions November 2001...

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Course 1 Solutions November 2001 Exams

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Course 1 Solutions 1 November 2001 1. A () () [] 12 For 1, 2, let event that a red ball is drawn form urn event that a blue ball is drawn from urn . Then if is the number of blue balls in urn 2, 0.44 Pr[ ] Pr[ ] Pr i i i Ri Bi x R RB B R R B B = = = == + = ∩∪∩ [][] [][] Pr Pr Pr Pr 41 6 6 10 16 10 16 Therefore, 32 3 3 32 2.2 16 16 16 2.2 35.2 3 32 0.8 3.2 4 RR BB x xx x x x +  =+  ++  + =+= + += + = = 2. C We are given that , 6 and 2 fxyd A d A ∫∫ It follows that () () 4, 2 4 , 2 4 6 2 2 20 R f xy dA f xydA −=   =−=
Course 1 Solutions 2 November 2001 3. A Since 34 5 2 175 SL A = , we have 4 1 5 2 2 4 1 5 2 2 31 5 2 2 6 3 5 2 2 262.5 0 for 0 , 0 131.25 0 for 0 , 0 140 0 for 0 , 0 28 0 for 0 , 0 S LA L A L S L A L S L A A S L A A => > > > > > > =− < > > It follows that S increases at an increasing rate as L increases, while S increases at a decreasing rate as A increases. 4. B Apply Baye’s Formula: [] () ( ) Pr Seri. Surv. Pr Surv. Seri. Pr Seri. Pr Surv. Crit. Pr Crit. Pr Surv. Seri. Pr Seri. Pr Surv. Stab. Pr Stab. 0.9 0.3 0.29 0.6 0.1 0.9 0.3 0.99 0.6   = + + == ++

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Course 1 Solutions 3 November 2001 5. A Let us first determine K . Observe that 1 1 1 1 60 30 20 15 12 137 11 2345 6 0 6 0 60 137 KK K K ++++  =+ + + + = =   = It then follows that [] () Pr Pr Insured Suffers a Loss Pr Insured Suffers a Loss 60 3 0.05 , 1,. ..,5 137 137 Nn N NN == =  = Now because of the deductible of 2, the net annual premium P EX = where 0 , if 2 2 , if 2 N X = −> Then, 5 3 313 3 2 1 2 3 0.0314 137 137 137 4 137 5 N PEX N N =  = + + =  6. E The line 543 yx −= has slope 4 5 . It follows that we need to find t such that 4 2 21 5 841 0 2 4 2 dy dy dx t dt dt t dx tt t t = + += = =
Course 1 Solutions 4 November 2001 7. A () ( ) ( ) ( ) ( ) () () 12 2 22 2 Cov , Cov , 1.2 Cov , Cov , Cov ,1.2 Cov Y,1.2Y Var Cov , 1.2Cov , 1.2Var Var 2.2Cov , 1.2Var Var 27.4 5 2.4 Var CC X Y X Y XX YX X Y Y X Y Y Y Y XEX EX YE Y =+ + + + + + + =− = = = ( ) ( ) ( ) 2 2 51.4 7 2.4 Var Var Var 2Cov , 1 Cov , Var Var Var 2 1 8 2.4 2.4 1.6 2 Cov , 2.4 2.2 1.6 1.2 2.4 8.8 EY XY X Y X Y X Y −= = += + + + =

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Course 1 Solutions 5 November 2001 7. Alternate solution: We are given the following information: [] 1 2 2 2 1.2 5 27.4 7 51.4 Var 8 CX Y Y EX EY XY =+ =  =  = = += Now we want to calculate () ( ) ( ) [ ] [] [] [] ()( ) 12 22 Cov , Cov , 1.2 1.2 1.2 2.2 1.2 1.2 2.2 1.2 5 7 5 1.2 7 27. CC X YX Y EXYX Y EXYEX Y E XX Y Y E X E Y E X E Y Y + +− + + =++−+ + + + + = i [] () ( ) 4 2.2 1.2 51.4 12 13.4 2.2 71.72 Y Y ++ =− Therefore, we need to calculate E XY first. To this end, observe []() [] 2 2 2 2 8V a r 2 2 5 7 27.4 2 51.4 144 2 65.2 XY EXY EXY E Y Y E X E Y Y EY Y Y Y = + + + −+ + + 8 65.2 2 36.6 = Finally, () () 1, 2 Cov 2.2 36.6 71.72 8.8 =
Course 1 Solutions 6 November 2001 8. D The function () , 1 7 Ft t ≤≤ , achieves its minimum value at one of the endpoints of the interval 17 t or at t such that () ( ) 01 ttt F tet e e t −−− == = Since 0 for 1 F tt <> , we see that F t is a decreasing function on the interval t <≤ . We conclude that 7 7 7 0.0064 Fe is the minimum value of F .

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Course1solutions_1101 - Course 1 Solutions November 2001...

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