PQ-6-Answer-331-2011-F

PQ-6-Answer-331-2011-F - Answers to Practice Questions 6...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Answers to Practice Questions 6 – ACTSC 331, Fall 2011 1. (a) The expectation is E [ T x ] = R ∞ t p ( τ ) x dt = 50 . (b) E ( K x ) = ∑ ∞ k =1 k p ( τ ) x = ∑ 100 k =1 100- k 100 = 49 . 5. (c) The probability is Pr { 10 ≤ T x < 11 } = 10 p ( τ ) x- 11 p ( τ ) x = 1 / 100 . (d) The probability is Pr { 10 ≤ T x < 11 , J x = 2 } = R 11 10 t p ( τ ) x μ (2) x ( t ) dt = 1 / 300. (e) The probability is f J | T (1 | 10) = μ (1) x (10) μ ( τ ) x (10) = 1 / 6 . (f) E [ T x | J x = 2] = R ∞ t f T,J ( t, 2) f J (2) dt = R 100 t/ 100 dt = 50. 2. We are given that p ( τ ) 60 = p (1) 60 p (2) 60 = 0 . 6375 and l ( τ ) 61 = l ( τ ) 60 p ( τ ) 60 = 6375 . Thus, p ( τ ) 61 = l ( τ ) 62 l ( τ ) 61 = 5538 6375 and q ( τ ) 61 = 1- p ( τ ) 61 = 0 . 1313 . 3. We wish to find q (2) 30 . We are given by (a) that 3 q (1) 30 = 2 q (2) 30 and by (b) that 0 . 25 = q (1) 30 q (2) 30 = 2 3 h q (2) 30 i 2 . Thus, q (2) 30 = 0 . 6124 ....
View Full Document

This note was uploaded on 01/04/2012 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.

Ask a homework question - tutors are online