sec_4_supp_sol

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

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Unformatted text preview: 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 , . 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 = . 9503 c. . 95 = P ( X > 135) = P X- 137 . 2 > 135- 137 . 2 = P ( Z >- 2 . 2 / ) - 2 . 2 / =- 1 . 645 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 > . 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- . 00 c. For a standard normal Z , we can find c such that . 90 = P (- c < Z < c ) which is the same as solving P ( Z > c ) = 0...
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sec_4_supp_sol - APPM 4/5570 Solutions to Problems from the...

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