Homework4 - Y (adding a constant doesn’t change the...

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HOMEWORK PROBLEMS #4 SOLUTIONS TO EVEN PROBLEMS 2. y - 1 1 2 P ( Y = y ) 0.5 0.25 0.25 4. Let A = valve 1 opens, B = valve 2 opens, and C = valve 3 opens. Note that { Y = 0 } = A ( B C ) { Y = 1 } = A ( B C ) - A B C { Y = 2 } = A B C Moreover, P ( A B C ) = P ( A ) P ( B ) P ( C ) = 0 . 8 3 = 0 . 512 P ( A ( B C )) = P ( A ) + P ( B C ) - P ( A B C ) = 0 . 8 + 0 . 8 2 - 0 . 8 3 = 0 . 928 y 0 1 2 P ( Y = y ) 0.072 0.416 0 . 512 12. E ( Y ) = 1( . 4) + 2( . 3) + 3( . 2) + 4( . 1) = 2 E ± 1 Y ² = 1( . 4) + 1 2 ( . 3) + 1 3 ( . 2) + 1 4 ( . 1) = 77 120 0 . 64 E ( Y 2 - 1) = E ( Y 2 ) - 1 = 1 2 ( . 4) + 2 2 ( . 3) + 3 2 ( . 2) + 4 2 ( . 1) - 1 = 5 - 1 = 4 Var( Y ) = E ( Y 2 ) - [ E ( Y )] 2 = 5 - 2 2 = 1 14. (a) μ = 7 . 9 (b) E ( Y 2 ) = 67 . 14, σ 2 = E ( Y 2 ) - μ 2 = 67 . 14 - 7 . 9 2 = 4 . 73, σ = 4 . 73 2 . 17 (c) μ ± 2 σ = 7 . 9 ± 2 4 . 73 = (3 . 55 , 12 . 25). Therefore, P ( μ - 2 σ Y μ + 2 σ ) = P (3 . 55 Y 12 . 25) = P (4 Y 12) = 1 - ( . 03 + . 01) = . 96 30. (Optional: extra credit) (a) Larger, since X is always larger than Y . (b) E ( X ) = E ( Y + 1) = E ( Y ) + E (1) = E ( Y ) + 1 = μ + 1. Yes this result agrees with my answer to (a). (c) Equal, since the spread in X will directly result from the spread of
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Unformatted text preview: Y (adding a constant doesn’t change the spread, it just moves the distribution to the left of to the right). (d) Var( X ) = E [( X-E ( X )) 2 ] = E [( Y + 1-( μ + 1)) 2 ] = E [( Y-μ ) 2 ] = Var( Y ) 34. Let X = daily cost for use of the tool. Then X = ($10) Y . From the given probability distribution we have that E ( Y ) = 1 . 3 and E ( Y 2 ) = 2 . 1. Therefore, Var( Y ) = 2 . 1-1 . 3 2 = 0 . 41. Finally (the units for the variance are ignored, since they don’t make sense), E ( X ) = ($10) E ( Y ) = $13, and Var( X ) = 10 2 Var( Y ) = 41 1...
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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