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Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 4.3 35. X ∼ N (8 . 8 , 2 . 8 2 ), ie: μ = 8 . 8, σ 2 = 2 . 8 2 , and σ = 2 . 8. a. P ( X ≥ 10) = P X 8 . 8 2 . 8 ≥ 10 8 . 8 2 . 8 ≈ P ( Z ≥ . 43) = 1 Φ(0 . 43) = Φ( . 43) = 0 . 3336 P ( X > 10) = P ( X ≥ 10) = 0 . 3336 b. P ( X > 20) = P X 8 . 8 2 . 8 > 20 8 . 8 2 . 8 ≈ P ( Z > 4) ≈ 0 given the accuracy of Table A. 3. c. P (5 ≤ X ≤ 10) = P 5 8 . 8 2 . 8 X 8 . 8 2 . 8 ≥ 10 8 . 8 2 . 8 ≈ P ( 1 . 36 ≤ Z ≤ . 43) = Φ(0 . 43) Φ( 1 . 36) = 0 . 6664 . 0869 = 0 . 5795 d. Want to solve P (8 . 8 c < X < 8 . 8 + c ) = 0 . 98 for c . Standardizing, this gives P 8 . 8 c 8 . 8 2 . 8 < X 8 . 8 2 . 8 < 8 . 8+ c 8 . 8 2 . 8 = P c 2 . 8 < Z < c 2 . 8 . By symmetry about zero of the zcurve, we want to find c so that P Z < c 2 . 8 = 0 . 98 + 0 . 01 = 0 . 99 . A reverse lookup in Table A. 3 gives us that c 2 . 8 ≈ 2 . 33 ....
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 '08
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 Normal Distribution, Probability theory, chloride concentration values

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