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Asst7-suggested - Statistics 427 Homework#7 Solutions to...

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Statistics 427: Homework # 7 Solutions to Suggested Problems 4.29 a. .9838 is found in the 2.1 row and the .04 column of the standard normal table so c = 2.14. b. P(0 Z c) = .291 Φ (c) = .7910 c = .81 c. P(c Z) = .121 1 – P(c Z) = P(Z < c) = Φ (c) = 1 – .121 = .8790 c = 1.17 d. P(–c Z c) = Φ (c) – Φ (–c) = Φ (c) – (1 – Φ (c)) = 2 Φ (c) – 1 Φ (c) = .9920 c = .97 e. P( c | Z | ) = .016 1 – .016 = .9840 = 1 – P(c | Z | ) = P( | Z | < c ) = P(–c < Z < c) = Φ (c) – Φ (–c) = 2 Φ (c) – 1 Φ (c) = .9920 c = 2.41 4.33 a. P(X 18) = 25 . 1 15 18 Z P = P(Z 2.4) = Φ (2.4) = .9452 b. P(10 X 12) = P(–4.00 Z –2.40) P(Z –2.40) = Φ (–2.40) = .0082 c. P( |X – 15| 1.5(1.25) ) = P( |Z| 1.5) = P(–1.5 Z 1.5) = Φ (1.5) – Φ (–1.5) = .8664 4.39 a. μ + σ⋅ (91 st percentile from std normal) = 30 + 5(1.34) = 36.7 b. 30 + 5( –1.555) = 22.225 c. μ = 3.000 μ m; σ = 0.140. We desire the 90 th percentile: 30 + 1.28(0.14) = 3.179 4.44 a. P( μ – 1.5 σ X μ + 1.5 σ ) = P(–1.5 Z 1.5) = Φ (1.50) – Φ (–1.50) = .8664 b. P( X < μ – 2.5 σ or X > μ + 2.5 σ ) = 1 – P( μ – 2.5 σ X μ + 2.5 σ ) = 1 – P(–2.5 Z 2.5) = 1 – .9876 = .0124 c. P( μ – 2 σ X μ σ or μ + σ X
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