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hwk3_sol

# hwk3_sol - ISOM 111 L6-L7 Fall 2009 1 Homework 3 Solutions...

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ISOM 111 L6-L7, Fall 2009 1 Homework 3 Solutions I . Problems 5.14 (p.194) of the textbook. Solution : (a) The probability curve f ( x ) is f ( x ) = ( 1 / 3 , if 3 x 6 0 , otherwise . (b) 2 3 4 5 6 7 0.0 0.1 0.2 0.3 0.4 Probability Curve of Rainfall x f(x) (c) May rainfall will be at least four inches with probability (6 - 4) / (6 - 3) = 2 / 3; at least five inches with probability (6 - 5) / (6 - 3) = 1 / 3; and at most 4.5 inches with probability (4 . 5 - 3) / (6 - 3) = 1 / 2. II . Problem 5.28[c(1), c(3), d] (p.209) of the textbook. Solution : (c) Let X be the IQ score of a randomly selected person. X is normally distributed with mean 100 and standard deviation 16. c(1) P ( X > 140) = P ( Z > (140 - 100) / 16) = P ( Z > 2 . 5) 0 . 5 - 0 . 4938 = 0 . 0062 .

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ISOM 111 L6-L7, Fall 2009 2 c(3) P (72 X 128) = P ((72 - 100) / 16 Z (128 - 100) / 16) = P ( - 1 . 75 Z 1 . 75) = 2 P (0 Z 1 . 75) 2 × 0 . 4599 = 0 . 9198 . (d) The percentage equals P ( X > 136) = P ( Z > (136 - 100) / 16) = P ( Z > 2 . 25) 0 . 5 - 0 . 4878 = 0 . 0122 = 1 . 22% . III . Problem 5.40 (p.211) of the textbook. Solution : Let X be the yearly health care expense of a randomly selected family of four. X is normally distributed with mean 3000 and standard deviation 500. We want to find x such that P ( X x ) = 0 . 33 , i.e., P ( Z ( x - 3000) / 500) = 0 . 33 . So the z -value of x is - z 0 . 33 ≈ - 0 . 44, and x = 500 z + 3000 \$2780 . IV . Problem 5.48 (p.215) of the textbook.
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hwk3_sol - ISOM 111 L6-L7 Fall 2009 1 Homework 3 Solutions...

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