November 2001 - Course 1 SOA Solutions

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7. Alternate solution: We are given the following information: C1 = X + Y C2 = X + 1.2Y E X = 5 [ ] 2 ⎡ ⎤ = E⎣X ⎦ EY = 7 [ ] 2 ⎡ ⎤ = 27.4 E⎣Y ⎦ 51.4 Var[ X + Y] = 8 Now we want to calculate Cov C , C = Cov X + Y, X + 1.2Y ( 1 2) ( ) = E⎡⎣( X + Y)( X + 1.2Y) ⎤⎦− E[ X + Y] iE[ X + 1.2Y] 2 2 = ⎡ ⎣ + + ⎤ ⎦− [ ] + [ ] + 2 2 = E⎡ ⎣X ⎤ ⎦+ 2.2E[ XY] + 1.2E⎡ ⎣Y ⎤ ⎦− ( 5 + 7) 5 + ( 1.2) 7 = 27. 4 + 2.2E[ XY] + 1.2( 51.4) −( 12)( 13.4) = 2.2E[ XY] −71.72 Therefore, we need to calculate E [ XY ] first. To this end, observe 2 2 8= Var[ X + Y] = E⎡( X + Y) ⎤− ( E[ X + Y] ) [ ] E XY ( )( [ ] [ ]) ( ) E X 2.2XY 1.2Y E X E Y E X 1.2E Y ⎣ ⎦ 2 2 2 = E ⎡ ⎣X + 2XY + Y ⎤ ⎦− ( E[ X] + E[ Y] ) 2 2 2 = E⎡ ⎣X ⎤ ⎦+ 2E[ XY] + E⎡ ⎣Y ⎤ ⎦− ( 5 + 7) = 27.4 + 2E[ XY] + 51.4 −144 = 2E[ XY] −65.2 ( 8 65.2) 2 36.6 = + = Finally, Cov CC = 2.2 36.6 − 71.72 = 8.8 ( 1, 2 ) ( ) Course 1 Solutions 5 November 2001
8. D The function F() t , 1≤t ≤ 7 , achieves its minimum value at one of the endpoints of the interval 1≤t ≤ 7 or at t such that −t −t −t 0= F′ () t = e − te = e ( 1−t) Since F′ () t < 0 for t > 1 , we see that F () t is a decreasing function on the interval 1< t ≤ 7 . We conclude that F 7 = 7e − = 0.0064 ( ) 7 is the minimum value of F. 9. D Let C = event that patient visits a chiropractor T = event that patient visits a physical therapist We are given that Pr[ C] = Pr[ T] + 0.14 Pr C∩T = 0.22 ( ) c c ( C ∩T ) Pr = 0.12 Therefore, c c 0.88 = 1− Pr ⎡ ⎣C ∩T ⎤ ⎦ = Pr C∪T = Pr C + Pr T −Pr C∩T = Pr[ T] + 0.14 + Pr[ T] −0.22 = 2 Pr T −0.08 or [ ] Pr[ T ] = ( 0.88 + 0.08) 2 = 0.48 [ ] [ ] [ ] [ ] Course 1 Solutions 6 November 2001
Page 1 and 2: Course 1 Solutions November 2001 Ex
Page 3 and 4: 3. A Since 3 4 2 5 S = 175L A , we
Page 5: 7. A Cov C , C = Cov X + Y, X + 1.2
Page 9 and 10: 12. E Observe that x f′ x = f′
Page 11 and 12: 17. B Note that 20 [ > ] = ( − )
Page 13 and 14: 21. D Let N1 and N 2 denote the num
Page 15 and 16: 24. D 1 1 4 2 100 x y must be maxim
Page 17 and 18: 27. D Let X denote the number of em
Page 19 and 20: 29. B Let X and Y denote repair cos
Page 21 and 22: 31. E The general solution of the d
Page 23 and 24: 35. E If X is the random variable r
Page 25: 39. D We are given that P x =− x

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