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Hmwk5Sol

# Hmwk5Sol - ISyE 2027 R D Foley Probability with...

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ISyE 2027 Probability with Applications Fall 2011 R. D. Foley Solution to Homework 5 October 19, 2011 1. Solution: ( 1 - 3 n ) n - 1 = ( 1 - 3 n ) n * ( 1 - 3 n ) (- 1 ) , the first part converges to e - 3 and the 2nd part converges to 1, so lim n ( 1 - 3 n ) n - 1 = e - 3 . 2. Solution: (a)Bernoulli. (b)Geometric. (c)Binomial. (d)Bernoulli. (e)Geometric. (f)None of the three. 3. Solution: (a) Pr { Y = k } = 1 3 , k = 0 1 2 , k = 1 1 6 , k = 2 (b)E [ Y ] = 1 3 × 0 + 1 2 × 1 + 1 6 × 2 = 1 2 + 1 3 = 5 6 . (c)E [ Y 2 ] = 1 3 × 0 2 + 1 2 × 1 2 + 1 6 × 2 2 = 1 2 + 2 3 = 7 6 . (d) Var ( Y ) = E [ Y 2 ] - ( E [ Y ]) 2 = 7 6 - 25 36 = 17 36 . (e) p Var ( Y ) = 17 6 . (f) c < E [ Y ] , so c can be 5 6 , you would still be willing play. 4. Solution: (a) ( 50 20 ) ( 100 40 ) . (b)1 - ( 50 40 )( 2 1 ) 40 ( 100 40 ) . (c) ( 50 k )( 2 2 ) k ( 50 - k 40 - 2 k )( 2 1 ) 40 - 2 k ( 100 40 ) . 5. Solution: (a) μ = E [ X ] = 1 3 × 9 10 + 7 × 1 10 = 1. (b)E [ X 2 ] = 1 9 × 9 10 + 49 × 1 10 = 5, σ 2 = E [ X 2 ] - ( E [ X ]) 2 = 5 - 1 = 4, so σ = 2. (c)Pr { X > μ + 3 σ } = Pr { X > 1 + 3 × 2 } = Pr { X > 7 } = 1 10 . (d)No. (e)Pr {| X - μ | > 3 σ } = Pr {| X - 1 | > 6 } = Pr { X > 7 } = 1 10 . (f)Pr { X > 7 } 6 E [ X ] 7 = 1 7 . (g)Pr {| X - μ | > 3 σ } 6 Var ( X ) ( 3 σ ) 2 = σ 2 9 σ 2 = 1 9 . 6. Solution: Suppose k = 1 1 k 2 = 1 c , then let Pr { X = k } = c k 2 , for k = 1, 2, 3, · · · .

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Hmwk5Sol - ISyE 2027 R D Foley Probability with...

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