Course 1 May 2000 Multiple Choice Exams - Society of Actuaries

More documents

Recommendations

Info

1. Answer: A Let R = event of referral to a specialist L = event of lab work We want to find P[R«L] = P[R] + P[L] – P[R»L] = P[R] + P[L] – 1 + P[~(R»L)] = P[R] + P[L] – 1 + P[~R«~L] = 0.30 + 0.40 – 1 + 0.35 = 0.05 . 2. Answer: D Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or 1999 . PA [ 1997] P[1997] P[1997|A] = PA [ 1997][ P[1997] + PA [ 1998] P[1998] + PA [ 1999] P[1999] (0.05)(0.16) = = 0.45 . (0.05)(0.16) + (0.02)(0.18) + (0.03)(0.20) 3. Answer: D Let T denote printer lifetime. Then f(t) = ½ e –t/2 , 0 £ t £ ∞ Note that 1 1 −t/2 −t/2 1 P[T £ 1] = ∫ e dt = e = 1 – e –1/2 = 0.393 2 0 0 P[1 £ T £ 2] = 2 1 −t/2 −t/2 2 ∫ e dt = e 1 = e –1/2 - e –1 = 0.239 2 1 Next, denote refunds for the 100 printers sold by independent and identically distributed random ⎧200 with probability 0.393 ⎪ variables Y 1 , . . . , Y 100 where Y i = ⎨100 with probability 0.239 i = 1, . . . , 100 ⎪⎩ 0 with probability 0.368 Now E[Y i ] = 200(0.393) + 100(0.239) = 102.56 Therefore, Expected Refunds = EY [ ] 100 ∑ i = 100(102.56) = 10,256 . i= 1 Course 1 2 May 2000
4. Answer: E First note R = 10/T . Then ⎡10 ⎤ ⎡ 10⎤ ⎛10 ⎞ F R (r) = P[R £ r] = P ⎢ ≤ r P T 1 FT T ⎥ = ⎢ ≥ = − ⎜ ⎟ r ⎥ . Differentiating with respect to r ⎣ ⎦ ⎣ ⎦ ⎝ r ⎠ f R (r) = F¢ R (r) = d/dr ⎛ ⎛10 ⎞ d 10 1 F ⎞ ⎛ ⎞⎛ T FT () t − ⎞ ⎜ − ⎜ ⎟ =− 2 r ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎝ ⎠⎠ ⎝dt ⎠⎝ r ⎠ d 1 F T() t = f T() t = since T is uniformly distributed on [8, 12] . dt 4 −1⎛−10⎞ 5 Therefore f R (r) = ⎜ 2 ⎟ = . 2 4 ⎝ r ⎠ 2r 5. Answer: C We are given f(t 1 , t 2 ) = 2/L 2 , 0 £ t 1 £ t 2 £ L . Therefore, E[T 1 2 + T 2 2 ] = = 2 4 2 ⎡ ⎤ 2 L L 4 L 3 t2 2 2∫ t2 dt2 = 2 ⎢ ⎥ = L 0 3 L ⎣ 3 ⎦ 3 0 L t2 2 2 2 ∫∫ ( t1 + t2 ) dt 2 1dt2 = L 0 0 3 t2 3 2 ⎧L L ⎪ ⎡t1 2 ⎤ ⎫⎪ 2 ⎧⎪ ⎛t2 3⎞ ⎫⎪ t 2 ⎨ 2 t1 dt1 t 2 2 dt2 L ∫⎢ + ⎥ ⎬= ⎨ ⎜ + ⎟ 3 L ∫ ⎬ ⎪⎩ 3 0⎣ ⎦0 ⎪⎭ ⎪⎩0⎝ ⎠ ⎪⎭ t 2 ( L, L) t 1 6. Answer: D Let X 1 and X 2 denote the measurement errors of the less and more accurate instruments, respectively. If N(m,s) denotes a normal random variable with mean m and standard deviation s, then we are given X 1 is N(0, 0.0056h), X 2 is N(0, 0.0044h) and X 1 , X 2 are independent. It 2 2 2 2 X1+ X2 0.0056 h + 0.0044 h follows that Y = is N (0, ) = N(0, 0.00356h) . Therefore, 2 4 P[-0.005h £ Y £ 0.005h] = P[Y £ 0.005h] – P[Y £ -0.005h] = P[Y £ 0.005h] – P[Y ≥ 0.005h] ⎡ 0.005h ⎤ = 2P[Y £ 0.005h] – 1 = 2P ⎢ Z ≤ ⎣ 0.00356h ⎥ - 1 = 2P[Z £ 1.4] – 1 = 2(0.9192) – 1 = 0.84 . ⎦ Course 1 3 May 2000
Page 1 and 2: SOCIETY OF ACTUARIES Mathematical F
Page 3 and 4: 1. The probability that a visit to
Page 5 and 6: 3. The lifetime of a printer costin
Page 7 and 8: 5. Let T 1 and T 2 represent the li
Page 9 and 10: 7. An insurance company’s monthly
Page 11 and 12: 9. The total claim amount for a hea
Page 13 and 14: 11. A company offers a basic life i
Page 15 and 16: 13. For a certain product priced at
Page 17 and 18: 15. In a certain town, the rate of
Page 19 and 20: 17. An insurance company has 120,00
Page 21 and 22: 19. In an analysis of healthcare da
Page 23 and 24: 21. A ball rolls along the polar cu
Page 25 and 26: 23. An insurance policy is written
Page 27 and 28: 25. An advertising executive claims
Page 29 and 30: 27. A car dealership sells 0, 1, or
Page 31 and 32: 29. An investor buys one share of s
Page 33 and 34: 31. Let R be the region bounded by
Page 35 and 36: 33. A blood test indicates the pres
Page 37 and 38: 35. A company insures homes in thre
Page 39 and 40: 37. Let S be the surface described
Page 41 and 42: 39. An insurance policy is written
Page 43: Course 1 May 2000 Answer Key 1. A 2
Page 47 and 48: 10. Answer: C Let T 1 be the time u
Page 49 and 50: 17. Answer: C The company’s spend
Page 51: 21. Answer: B Denote arc length by
Page 54 and 55: 33. Answer: B Let Y = positive test

Course 1 May 2000 Multiple Choice Exams - Society of Actuaries

Create successful ePaper yourself

Delete template?

Save as template?