EXAM P SAMPLE SOLUTIONS

More documents

Recommendations

Info

$KSU Math Circle Summer Camp Application Form.pdf$

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

or more precisely, ( ) g y ( ) 12 ⎧ 32 ⎪15y 1 − y , 0 < y< 1 = ⎨ ⎪⎩ 0 otherwise -------------------------------------------------------------------------------------------------------- 119. Solution: D The diagram below illustrates the domain of the joint density We are told that the marginal density function of X is while yx It follows that Therefore, ( ) = 1, < < + 1 f yx x y x x f ( xy , ) of Xand Y. ( ) = 1,0< < 1 f x x ⎧1 if 0< x < 1, x< y< x+ 1 f ( x, y) = fx( x) fyx( y x) =⎨ ⎩0 otherwise [ ] [ ] 1 1 2 2 ∫ ∫ Pr Y > 0.5 = 1−Pr Y ≤ 0.5 = 1− dydx 0 1 1 1 2 2 1 2 ⎛1 ⎞ ⎡1 1 2 ⎤ 1 1 7 2 = 1− ∫ y 1 1 0 x dx = −∫ x dx x x 0 ⎜ − ⎟ = − − 0 = 1− + = ⎝2 ⎠ ⎣ ⎢2 2 ⎦ ⎥ 4 8 8 [Note since the density is constant over the shaded parallelogram in the figure the solution is also obtained as the ratio of the area of the portion of the parallelogram above y = 0.5 to the entire shaded area.] Page 52 of 55 x
120. Solution: A We are given that X denotes loss. In addition, denote the time required to process a claim by T. ⎧3 2 1 3 ⎪ x ⋅ = x, x< t < 2 x,0≤ x≤ 2 Then the joint pdf of X and T is f( x, t) = ⎨8 x 8 ⎪⎩ 0, otherwise. Now we can find P[T ≥ 3] = 4 2 4 2 3 ⎡ 3 2⎤ ∫∫ xdxdt = 8 ∫⎢ x 16 3 t /2 3⎣ ⎥ ⎦t/2 = 11/64 = 0.17 . 4 4 ⎛12 3 2⎞ ⎡1213⎤ dt = ∫⎜ − t ⎟dt= − t 16 64 16 64 3⎝ ⎠ ⎣ ⎢ ⎦ ⎥ 3 12 ⎛36 27 ⎞ = −1−⎜ − ⎟ 4 ⎝16 64 ⎠ t t=2x 4 3 2 1 1 2 t=x x -------------------------------------------------------------------------------------------------------- 121. Solution: C The marginal density of X is given by 1 3 1 1 1 1 2 ⎛ xy ⎞ ⎛ x⎞ fx( x) = ∫ ( 10 − xy ) dy = ⎜10y− ⎟ = 10 0 ⎜ − ⎟ 64 64 ⎝ 3 ⎠ 64 ⎝ 3 ⎠ Then 2 10 10 1 ⎛ x ⎞ ∫2 x ∫ ⎜ ⎟ x 2 E( X) = xf ( x) dx= 10x− d 64 ⎝ 3 ⎠ = 1 1000 8 ⎡⎛ ⎞ ⎛ ⎞⎤ 500 20 64 ⎢⎜ − ⎟−⎜ − ⎟ 9 9 ⎥ = 5.778 ⎣⎝ ⎠ ⎝ ⎠ ⎦ 0 Page 53 of 55 10 3 ⎛ 2 x ⎞ 1 = ⎜5x−⎟ 64 ⎝ 9 ⎠ 2
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 and 34: ( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1
Page 35 and 36: 40 − 3n −2≥ n To find such an
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: 116. Solution: D Denote the number
Page 55: 124. Solution: A − x −2 y Becau

EXAM P SAMPLE SOLUTIONS

Create successful ePaper yourself

Delete template?

Save as template?