EXAM P SAMPLE SOLUTIONS

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$MATH 2203 @Exam 4 (Version 1) Solutions November 9, 2005 S. F. ...$

-------------------------------------------------------------------------------------------------------- 81. Solution: C 1 Let X 1, . . . , X25 denote the 25 collision claims, and let X = (X 1 + . . . +X25) . We are 25 given that each Xi (i = 1, . . . , 25) follows a normal distribution with mean 19,400 and standard deviation 5000 . As a result X also follows a normal distribution with mean 19,400 and standard deviation 1 (5000) = 1000 . We conclude that P[ X > 20,000] 25 ⎡X −19,400 20,000 −19,400⎤ ⎡X −19,400 ⎤ = P⎢ > = P 0.6 1000 1000 ⎥ ⎢ > ⎣ ⎦ ⎣ 1000 ⎥ = 1 − Φ(0.6) = 1 – 0.7257 ⎦ = 0.2743 . -------------------------------------------------------------------------------------------------------- 82. Solution: B Let X 1, . . . , X1250 be the number of claims filed by each of the 1250 policyholders. We are given that each Xi follows a Poisson distribution with mean 2 . It follows that E[X i] = Var[X i] = 2 . Now we are interested in the random variable S = X 1 + . . . + X1250 . Assuming that the random variables are independent, we may conclude that S has an approximate normal distribution with E[S] = Var[S] = (2)(1250) = 2500 . Therefore P[2450 < S < 2600] = ⎡2450 −2500 S−2500 2600 −2500⎤ ⎡ S−2500 ⎤ P⎢ < < ⎥ = P 1 2 2500 2500 2500 ⎢ − < < 50 ⎥ ⎣ ⎦ ⎣ ⎦ ⎡S−2500 ⎤ ⎡S−2500 ⎤ = P ⎢ < 2 − P 1] = P[Z < 2] + P[Z < 1] – 1 ≈ 0.9773 + 0.8413 – 1 = 0.8186 . -------------------------------------------------------------------------------------------------------- 83. Solution: B Let X1,…, Xn denote the life spans of the n light bulbs purchased. Since these random variables are independent and normally distributed with mean 3 and variance 1, the random variable S = X 1 + … + Xn is also normally distributed with mean μ = 3n and standard deviation σ = n Now we want to choose the smallest value for n such that ⎡ S−3n 40−3n⎤ 0.9772 ≤ Pr[ S > 40] = Pr ⎢ > ⎥ ⎣ n n ⎦ This implies that n should satisfy the following inequality: Page 34 of 55
40 − 3n −2≥ n To find such an n, let’s solve the corresponding equation for n: 40 − 3n − 2 = n − 2 n = 40−3n 3n−2 n − 40= 0 3 n+ 10 n −4 = 0 ( )( ) n = 4 n = 16 -------------------------------------------------------------------------------------------------------- 84. Solution: B Observe that E[ X + Y] = E[ X] + E[ Y] = 50 + 20 = 70 Var [ X + Y ] = Var [ X ] + Var [ Y ] + 2 Cov[ X , Y ] = 50 + 30 + 20 = 100 for a randomly selected person. It then follows from the Central Limit Theorem that T is approximately normal with mean ET [ ] = 100( 70) = 7000 and variance 2 Var [ T ] = 100( 100) = 100 Therefore, ⎡T−7000 7100 −7000⎤ Pr[ T < 7100] = Pr ⎢ < ⎣ 100 100 ⎥ ⎦ = Pr[ Z < 1] = 0.8413 where Z is a standard normal random variable. Page 35 of 55
Page 1 and 2: SOCIETY OF ACTUARIES/CASUALTY ACTUA
Page 3 and 4: 4. Solution: A For i = 1, 2, let Ri
Page 5 and 6: 9. Solution: B Let M = event that c
Page 7 and 8: 14. Solution: A 1 11 1 1 1 ⎛1⎞
Page 9 and 10: 19. Solution: B Apply Bayes’ Form
Page 11 and 12: 25. Solution: B Let Y = positive te
Page 13 and 14: 31. Solution: D Let X denote the nu
Page 15 and 16: 36. Solution: B To determine k, not
Page 17 and 18: 41. Solution: E Let X = number of g
Page 19 and 20: 44. Solution: C If k is the number
Page 21 and 22: 5 ( ) = ∑ ( ) Pr[ = ] E⎡⎣f X
Page 23 and 24: 54. Solution: B Let Y denote the cl
Page 25 and 26: 59. Solution: B The distribution fu
Page 27 and 28: 64. Solution: A Let X denote claim
Page 29 and 30: 68. Solution: C Note that X has an
Page 31 and 32: 74. Solution: E First note R = 10/T
Page 33: ( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1
Page 37 and 38: -----------------------------------
Page 39 and 40: 92. Solution: B Let X and Y denote
Page 41 and 42: Consequently, 1 2 1 2 ( , ) 34 f t
Page 43 and 44: 99. Solution: C We use the relation
Page 45 and 46: 105. Solution: A The calculation re
Page 47 and 48: 2 [ X Y] E⎡( X Y) ⎤ ( E[ X Y] )
Page 49 and 50: 111. Solution: E 3 f Pr ⎡ ⎣1< Y
Page 51 and 52: 116. Solution: D Denote the number
Page 53 and 54: 120. Solution: A We are given that
Page 55: 124. Solution: A − x −2 y Becau

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