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SECTION 10 Hence, fY ( y) = FY! ( y) = 1 2 y . Answer A. 24. An insurer has 10 independent one-year term life insurance policies. The face amount of each policy is 1000. The probability of a claim occurring in the year under consideration is 0.1. Find the probability that the insurer will pay more than the total expected claim for the year. A. 0.01 B. 0.10 C. 0.16 D. 0.26 E. 0.31 Solution. The expected claim from each individual policy is 0.10 !1000 = 100. Thus the overall expected claim is 1000. The total claim for the year will be more than this expected claim if there is more than 1 death. The number of claims N has the binomial distribution with n = 10, p = 0.10. We want the probability that N > 1, i.e., Pr N > 1 Answer D. ( ) = ( ) = 1! Pr( N = 0) + Pr( N = 1) " 10% = 1! # $ 0 & ' ( 0.100 ( 0.90 10 + 10 " % # $ 1 & ' ( 0.101 ( 0.90 9 " % # $ & ' = = 1! 0.90 10 + 0.90 9 ( ) = 1! 1.90 ( 0.90 9 ) 0.2639. 25. An insurance policy is being issued for a loss with the following discrete distribution: ! 2, with probability 0.4, X = " # 20, with probability 0.6. Your job as the actuary is to set up a deductible d for this policy so that the expected payment by the insurer is 6. Find the deductible. A. 1 B. 5 C. 7 D. 10 E. 15 Solution. The large loss of 20 will be paid after deductible as 20 – d, and this can happen with probability 0.60, so that the expected payment of that loss is 0.6 ! ( 20 " d) = 12 " 0.6d. This amount is equal to 6 if d = 10. With that large of a deductible, the smaller loss of 2 must always be absorbed by the insured, so it will not affect the expected payment, and therefore with the deductible of 10, the expected payment is 6. Any smaller deductible will increase the expected payment, and a larger deductible will decrease it. Thus the deductible must equal to 10. Answer D. ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 280 -
26. For a Poisson random variable N with mean ! find lim E( N N # 1). !"0 !"0 n=1 !"0 n=1 PRACTICE EXAMINATION NO. 6 A. ! Solution. B. 0 C. 1 D. e E. Cannot be determined n ! Let fN ( n) = n! " e#! for n = 0,1,2,… be the probability function of this random variable. We have +% +% fN ( n) lim E( N N # 1) = lim& n $ f n N # 1 !"0 N N #1 ( ) = lim& n $ = Pr N # 1 Answer C. = lim +% & !"0 n=1 ( ) ( ) fN n n $ 1' fN 0 1 = lim !"0 1' fN 0 ! = lim !"0 1' e '! ( ) +% & n=0 = 1 = lim !"0 1 ' fN 0 n $ f N n lim de l'Hospital !"0 ( ) = ( ) n $ fN ( n) = ( ) n=1 E( N ) 1 ' fN ( 0) = ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 281 - lim !"0 1 = 1. '! e 27. X is a normal random variable with mean zero and variance 1 and Y is distributed 2 exponentially with mean 1. X and Y are independent. Find the probability Pr Y > X 2 ( ). 1 e A. B. C. e ! Solution. We know that 1 fX ( x) = 1 2 2! e for !" < x < +", and ! y fY ( y) = e for y > 0, so that " 1 # 2 & % % 2 % $ % x ( ( 1 ( 2 ' ( fX,Y ( x, y) = 1 ! e" x2 " y e +% & 1 1 2 D. E. 2! 2 2 = 1 ! e" x2
Page 1 and 2: PRACTICE EXAMINATION NUMBER 6 1. An
Page 3 and 4: ! 1 $ A. # & " # ! % & B. ! 1 $ # &
Page 5 and 6: PRACTICE EXAMINATION NO. 6 17. The
Page 7 and 8: PRACTICE EXAMINATION NO. 6 27. X is
Page 9 and 10: We have: 40 " ( ) !2 1 = C 10 + x 0
Page 11 and 12: with density function ! 2x, for 0 <
Page 13 and 14: PRACTICE EXAMINATION NO. 6 ! 4 1.5
Page 15 and 16: PRACTICE EXAMINATION NO. 6 Let W =
Page 17 and 18: "# PRACTICE EXAMINATION NO. 6 The d
Page 19 and 20: 20 ! z 10 PRACTICE EXAMINATION NO.
Page 21 and 22: Similarly, Pr X( 7) < m PRACTICE EX
Page 23 and 24: This density at 1 ! 1$ , " # 2 2% &
Page 25 and 26: The survival function for the Weibu
Page 27 and 28: PRACTICE EXAMINATION NO. 6 21. Let
Page 29: PRACTICE EXAMINATION NO. 6 Let z0.6
Page 33: Find the variance of N. A. 0.512 B.

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