November 2001 - Course 1 SOA Solutions

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5. A Let us first determine K. Observe that ⎛ 1 1 1 1 ⎞ ⎛60 + 30 + 20 + 15 + 12 ⎞ ⎛137 ⎞ 1= K⎜1+ + + + ⎟= K⎜ ⎟= K⎜ ⎟ ⎝ 2 3 4 5 ⎠ ⎝ 60 ⎠ ⎝ 60 ⎠ 60 K = 137 It then follows that Pr N = n = Pr ⎣⎡ N = n Insured Suffers a Loss⎤⎦Pr Insured Suffers a Loss 60 3 = ( 0.05 ) = , N = 1,...,5 137N 137N P = E X where [ ] [ ] Now because of the deductible of 2, the net annual premium [ ] ⎧0 , if N ≤ 2 X = ⎨ ⎩N − 2 , if N > 2 Then, P E [ X 5 ] ( N 3 ⎛ 1 ⎞ ⎡ 3 ⎤ ⎡ 3 ⎤ = = ∑ − 2) = () 1 2 3 0.0314 N = 3 ⎜ ⎟+ ⎢ ⎥+ ⎢ ⎥ = 137N ⎝137 ⎠ ⎣137( 4) ⎦ ⎣137( 5) ⎦ 6. E The line 5y− 4x= 3 has slope 4 5 4 dy dy dx 2t = = = 5 dx dt dt 2t + 1 8t+ 4= 10t 2t = 4 t = 2 . It follows that we need to find t such that Course 1 Solutions 3 November 2001
7. A Cov C , C = Cov X + Y, X + 1.2Y ( 1 2) ( ) = Cov ( X, X) + Cov ( Y, X) + Cov ( X,1.2Y) + Cov( Y,1.2Y) = Var X + Cov ( X, Y) + 1.2Cov ( X, Y) + 1.2VarY = Var X + 2.2Cov ( X, Y) + 1.2VarY ( ) ( ( )) 2 ( ) ( E Y ) 2 2 2 Var X = E X − E X = 27.4 − 5 = 2.4 VarY 2 2 = E Y − ( ) = 51.4 − 7 = 2.4 ( X + Y) = X + Y + ( X Y) Var Var Var 2Cov , 1 Cov ( XY , ) = Var ( X+ Y) −Var X−VarY 2 1 = ( 8 − 2.4 − 2.4 ) = 1.6 2 Cov , = 2.4 + 2.2 1.6 + 1.2 2.4 = 8.8 ( C C ) ( ) ( ) 1 2 ( ) Course 1 Solutions 4 November 2001
Page 1 and 2: Course 1 Solutions November 2001 Ex
Page 3: 3. A Since 3 4 2 5 S = 175L A , we
Page 7 and 8: 8. D The function F() t , 1≤t ≤
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

November 2001 - Course 1 SOA Solutions

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