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# plugin-HW_10_Solutions - HW#10Solutions...

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HW #10 Solutions Ang and Tang: 6.2, 6.6, 6.9, 6.13 Due: Th 12/1/2011

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6.2 (a) Since x = 65, n = 50, σ = 6, a 2-sided 99% interval for the true mean is given by the limits 65 ± k 0.005 50 6 < μ > 0.99 = (65 – 2.58 × 50 6 , 65 + 2.58 × 50 6 ) = ( 62.81, 67.19 ) (in mph) (b) Requiring 2.58 n 6 1 n (2.58 × 6) 2 = 239.6304 n = 240 (240 – 50) = 190 additional vehicles must be observed. Before we proceed to (c) and (d), let J X and M X denote John and Mary’s sample means, respectively. Both are approximately normal with mean μ (unknown true mean speed) and standard deviation n 6 , hence their difference D = J X M X has an (approximate) normal distribution with mean = μ μ = 0, and standard deviation = 2 2 6 6 + n n = 6 n 2 , i.e. D ~ N(0, 6 n 2 ) Hence (c) When n = 10, P( J X M X > 2) = P(D > 2) = P( 5 / 6 0 2 > D D D σ μ ) = 1 – Φ (0.745) 0.228 (d) When n = 100, P(D > 2) = 1– Φ ( 50 / 6 0 2 ) = 1 – Φ ( 2.357) 0.0092
6.6 (a) N = 30, x = 12.5 tons Assume σ = 3 tons Assume σ = 3 tons < x μ > 0.99 = ( x + k 0.005 n σ , x + k 0.995 n ) = ( 12.5 – 2.58 × 30 3 , 12.5 + 2.58 × 30 3 ) = (12.5 – 1.413, 12.5 + 1.413) = (11.087, 13.913) tons (b)

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## This note was uploaded on 01/02/2012 for the course CIVIL ENG 408 at USC.

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plugin-HW_10_Solutions - HW#10Solutions...

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