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sec_4_supp_sol

# sec_4_supp_sol - APPM 4/5570 Solutions to Problems from the...

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APPM 4/5570 Solutions to Problems from the Supplementary Exercises at the End of Chapter 4 103. Let X be the contents of a single jar. Then X N (137 . 2 , 1 . 6 2 ). a. P ( X > 135) = P X - 137 . 2 1 . 6 > 135 - 137 . 2 1 . 6 = P ( Z > - 1 . 38) = 0 . 9162 b. Let Y be the number out of the ten jars that contain more than the stated contents. Then Y bin (10 , 0 . 9162). So, P ( Y 8) = P ( Y = 8) + P ( Y = 9) + P ( Y = 10) = 10 8 ! (0 . 9162) 8 (0 . 0838) 2 + 10 9 ! (0 . 9162) 9 (0 . 0838) + 10 10 ! (0 . 9162) 10 = 0 . 9503 c. 0 . 95 = P ( X > 135) = P X - 137 . 2 σ > 135 - 137 . 2 σ = P ( Z > - 2 . 2 ) - 2 . 2 = - 1 . 645 σ 1 . 34 105. Let X = grain size. Then X N (96 , 14 2 ). P ( X > 100) = P X - 96 14 > 100 - 96 14 = P ( Z > 0 . 29) = 0 . 3859 b. P (50 < X < 80) = P 50 - 96 14 < X - 96 14 < 80 - 96 14 = P ( - 3 . 29 < Z < - 1 . 14) = 0 . 1271 - 0 . 00

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c. For a standard normal Z , we can find c such that 0 . 90 = P ( - c < Z < c ) which is the same as solving P ( Z > c ) = 0 . 05. From the table we see that c = 1 . 645. So, 0 . 90 = P ( - 1 . 645 < Z < 1 . 645) = P - 1 . 645 < X - 96 14 < 1 . 645 = P (72 . 97 < X < 119 . 03) . The interval is (72 . 97 , 119 . 03). 111. a. For the normal or “bell curve”, the mode (highest point) is in the center at μ . b. No, the uniform distribution on the interval ( A, B ) does not have a single mode. The pdf is a flat line. Any point in ( A, B ) could be considered a mode.
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sec_4_supp_sol - APPM 4/5570 Solutions to Problems from the...

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