HW5_soln

# HW5_soln - AMS 410 Actuarial Mathematics Fall 2009 Homework...

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Unformatted text preview: AMS 410 Actuarial Mathematics Fall 2009 Homework 5 solution 12/8/2009 1. Normal . If for a certain normal random variable X , P ( X < 500) = 0 . 5 and P ( X > 650) = . 0227 , nd the standard deviation of X . Solution : Normal distribution is symmetric about the mean. Thus if X ∼ N ( μ,σ 2 ) , P ( X < μ ) = 0 . 5 . In this question, μ = 500 . P ( X > 650) = P ( X- 500 σ > 650- 500 σ ) = P ( Z > 150 σ ) = 1- Φ( 150 σ ) = 0 . 0227 . So Φ( 150 σ ) = 0 . 9773 . Check the Normal distribution table, . 9773 = Φ(2) . So the standard deviation of X is σ = 75 . 2. Normal . There are two independent random variables: W and X , both normally distributed with the same mean. The variances of W and X are 4 and 12 respectively. What is P ( | W- X | < 1) ? Solution : De ne Y = W- X , then Y ∼ N (0 ,σ 2 = 4+12 = 16) . P ( | W- X | < 1) = P ( | Y | < 1) = P (- 1 < Y < 1) = P (- 1- 4 < Y- 4 < 1- 4 ) = P (- 1 4 < Z < 1 4 ) = Φ( 1 4 )- Φ(- 1 4 ) = 2Φ( 1 4 )- 1 = 2 × . 5987- 1 = 0 . 1974...
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HW5_soln - AMS 410 Actuarial Mathematics Fall 2009 Homework...

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