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Sec_6_solutions_07

# Sec_6_solutions_07 - ln(1000 µ ML – µ VL ML 2 VL 2 =...

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CEE 304 – Section #6 Example Problems 1. The maximum annual 24-hour rainfall in Vermont has a mean of 2.5 inches and a standard deviation of 0.7 inches. The distribution is often described by a Gumbel distribution (the appropriate asymptotic distribution) or the lognormal distribution (a reasonable distribution for describing strictly positive variates). Estimate the 50-year (98 th percentile) rainfall depth using both distributions. How far can you get with the gamma distribution? Ans. Gumbel Distribution: x 0.98 = 4.315 in. Lognormal Distribution: x 0.98 = 4.238 in. Gamma Distribution: x 0.98 = 4.16 in. [computed using Excel] 2. For volume  V ~ Lognormal: ln V = VL  ~ N[ µ VL σ VL 2 ]  where one obtains  VL 2 = ln[ 1 + (0.10/0.50) 2  ] = 0.03922 = (0.20) 2 ; µ VL  = ln(0.5) – 0.5 VL 2   = – 0.7134      For mass  M ~ Lognormal: ln M = ML  ~ N[ µ ML ML 2 ]  where  µ ML   = ln(150) = 5.01; and  ML 2  = ln[ 1 + (0.20) 2  ] = (0.198) 2 Then ln(C) ~ N[ µ CL , CL 2 ] = N[
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Unformatted text preview: ln(1000) + µ ML – µ VL , ML 2 + VL 2 ] = N[–1.18, 0.078] Now: E[C] = exp[ µ CL + 0.5 CL 2 ] = 0.318 mg/l ; Coef. of Var. C = exp( CL 2 )-1 = 0.286 3. A structural engineer believes that the compressive strength C of some concrete columns supporting a very long bridge is distributed with a mean of 8000 psi and a standard deviation of 1500 psi. Compute c d , which is the strength exceeded by 95% of these columns, based on the following distributions: a.) Normal Distribution b.) Lognormal Distribution c.) Gamma Distribution d.) Gumbel Distribution e.) Weibull Distribution [with k = 6.22, Γ (1 + 1 / k ) = 0.9295] Ans . c d is the 5 th percentile (x 0.05 ) a.) Normal Distribution x 0.05 = 5532.5 psi b.) Lognormal Distribution x 0.05 = 5790.8 psi c.) Gamma Distribution x 0.05 = 5700 psi[using Excel] d.) Gumbel Distribution x 0.05 = 6041.7 psi e.) Weibull Distribution [with k = 6.22, Γ (1 + 1 / k ) = 0.9295] x 0.05 = 5339 psi...
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