plugin-HW_04_Solutions

plugin-HW_04_Solutions - HW #4 Solutions (Fall 2011)...

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HW #4 Solutions (Fall 2011) Assigned problems Ang & Tang: 3.38, 3.41, 3.42, 3.45, 3.46 3.38 Let, T = The present traffic volume = N(200,60) (a) P(T>350) = 1 - Φ ((350 - 200)/60) = 1 - Φ (2.5) = 0.00621 (b) Let, T 1 = Traffic volume after 10 years. μ T 1 = 200 + 0.10 × 200 × 10 = 400 δ T 1 = δ T = 60/200 = 0.3 Hence, σ T 1 = 0.3 × 400 = 120 So, T 1 is N(400, 120) P(T 1 >350) = 1 - Φ ((350 - 400)/120) = 1 - Φ (0.42) = 0.662 (c) C = Capacity of airport after 10 years. P(T 1 >C) = 0.00621 or, 1 - Φ ((C - 400)/120) = 1 – 0.99379 = 1- Φ (2.5) or, (C - 400)/120 = 2.5 or, C = 300 + 400 = 700
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3.41 (b) Let X be her cylinder’s strength in kips. To be the second place winner, X must be above 70 but below 100, hence P (70 < X < 100) 100 80 70 80 ( ) ( ) 20 20 (1) ( 0.5) 0.841 0.309 0.532 - - = Φ - Φ = Φ - Φ - = - = (c) P ( X > 100 | X > 90) = P ( X > 100 and X > 90) / P ( X > 90) = P ( X > 100) / P ( X > 90) = {1 – Φ [(100 – 80)/20]} / {1 – Φ [(90 – 80)/20]} = [1 – Φ (1)] / [1 –
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plugin-HW_04_Solutions - HW #4 Solutions (Fall 2011)...

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