November 2001 - Course 1 SOA Solutions

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19. B The amount of money the insurance company will have to pay is defined by the random variable ⎧1000 x if x< 2 Y = ⎨ ⎩2000 if x ≥ 2 where x is a Poisson random variable with mean 0.6 . The probability function for X is −0.6 k e ( 0.6) p( x) = k = 0,1, 2,3 $ and k! k −0.6 −0.6 ∞ 0.6 EY [ ] = 0 + 1000( 0.6) e + 2000e ∑ k = 2 k ! k −0.6 ⎛ −0.6 ∞ 0.6 −0.6 −0.6 ⎞ = 1000( 0.6) e + 2000⎜e ∑ −e −( 0.6) e k = 0 ⎟ ⎝ k! ⎠ k ∞ −0.6 ( 0.6) −0.6 −0.6 −0.6 −0.6 = 2000e ∑ −2000e − 1000( 0.6) e = 2000 −2000e −600e k = 0 k! = 573 k 2 2 −0.6 2 −0.6 ∞ 0.6 E⎡ ⎣Y ⎤ ⎦ = ( 1000) ( 0.6) e + ( 2000) e ∑ k = 2 k! k 2 −0.6 ∞ 0.6 2 −0.6 2 2 −0.6 = ( 2000) e ∑ −( 2000) e −⎡( 2000) −( 1000) ⎤( 0.6) e k = 0 k! ⎣ ⎦ 2 2 −0.6 2 2 = ( 2000) −( 2000) e −⎡ −0.6 ( 2000) − ( 1000) ⎤( 0.6) e ⎣ ⎦ = 816,893 Var 2 [ Y] = E⎡ ⎣Y ⎤ ⎦−{ E[ Y] } 2 = 816,893−( 573) = 488,564 [ Y ] Var = 699 2 20. B The graph of E′ tells us that employment increases from April through August (because E t E′ t ≤ then). ′() > 0 then) and does not increase from August through April (since () 0 It follows that employment is a minimum in April. Course 1 Solutions 11 November 2001
21. D Let N1 and N 2 denote the number of claims during weeks one and two, respectively. Then since N1 and N 2 are independent, 7 [ N1+ N2 = ] = ∑ [ N1 = n] [ N2 = −n] n= 0 Pr 7 Pr Pr 7 7 ⎛ 1 ⎞⎛ 1 ⎞ = ∑n= 0⎜ n+ 1⎟⎜ 8−n ⎟ ⎝2 ⎠⎝2 ⎠ 7 1 = ∑n= 0 9 2 8 1 1 = = = 9 6 2 2 64 22. D The amount of the drug present peaks at the instants an injection is administered. If A(n) is the amount of the drug remaining in the patient’s body at the time of the nth injection, then A 1 = 250 () −16 ( 2) = 250( 1+ e ) A % n−1 −k 6 An ( ) = 250∑ e k = 0 and the least upper bound is ∞ −k 6 250 lim An ( ) = 250∑ e = = 1628 (noting the n→∞ k = 0 −16 1− e −16 series is geometric with ratio e ) Course 1 Solutions 12 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 and 6: 7. A Cov C , C = Cov X + Y, X + 1.2
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: 17. B Note that 20 [ > ] = ( − )
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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