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Chapter_3_Problem_Solutions

# Chapter_3_Problem_Solutions - ACTSC 232 – Spring 2011...

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Unformatted text preview: ACTSC 232 – Spring 2011 Solutions to Chapter 3 Book Problems Exercise 3.2 Before we calculate the requested fractional probabilities, we first calculate some useful integer year probabilities: p 52 = ‘ 53 ‘ 52 = 89 , 089 89 , 948 = 0 . 99045 q 52 = 1- p 52 = 0 . 00955 2 p 53 = ‘ 55 ‘ 53 = 87 , 208 89 , 089 = 0 . 97889 5 p 53 = ‘ 58 ‘ 53 = 83 , 940 89 , 089 = 0 . 94220 p 54 = ‘ 55 ‘ 54 = 87 , 208 88 , 176 = 0 . 98902 q 54 = 1- p 54 = 0 . 01098 p 55 = ‘ 56 ‘ 55 = 86 , 181 87 , 208 = 0 . 98822 q 55 = 1- p 55 = 0 . 01178 2 p 56 = ‘ 58 ‘ 56 = 83 , 940 86 , 181 = 0 . 97400 p 58 = ‘ 59 ‘ 58 = 82 , 719 83 , 940 = 0 . 98545 q 58 = 1- p 58 = 0 . 01455 1 (a) We first note that . 6 p 52 = . 4 p 52 0 . 2 p 52 . 4 so that . 2 p 52 . 4 = . 6 p 52 . 4 p 52 and . 2 q 52 . 4 = 1- . 6 p 52 . 4 p 52 . 2 q 52 . 4 = 1- . 6 p 52 . 4 p 52 = 1- 1- . 6 q 52 1- . 4 q 52 UDD = 1- 1- (0 . 6) q 52 1- (0 . 4) q 52 UDD = 1- 1- (0 . 6)(0 . 00955) 1- (0 . 4)(0 . 00955) UDD = 0 . 00192 (b) We start as before, and then apply the Constant Force assumption: . 2 q 52 . 4 = 1- . 6 p 52 . 4 p 52 CF = 1- ( p 52 ) . 6 ( p 52 ) . 4 CF = 1- ( p 52 ) . 2 CF = 1...
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Chapter_3_Problem_Solutions - ACTSC 232 – Spring 2011...

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