Hw__7_Solutions_to_POST

# Hw__7_Solutions_to_POST - #1 Given E[ln(S = ln(40 ksi...

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#1 Given: E[ln(S)] = ln(40 ksi); Var[ln(S)] = (0.1)2 E[A] = 1.0 in 2 ; Var[A] = (0.05 in 2 ) 2 a) Assuming S and A are independent, find the distirubiton of the yield force: F = S * A F is lognormal because it is the product of two lognormal random variables. E[ln(F)] = E[ln(SA)] = E[ln(S) + ln(A)] = E[ln(S)] + E[ln(A)] Var[ln(F)] = Var[ln(S) + ln(A)] = Var[ln(S)] + Var[ln(A)] we have E[ln(S)] and Var[ln(S)], need to find E[ln(A)] and Var[ln(A)], Var[ln(A)] = ln{1 + Var[A]/(E[A]) 2 } = ln{1 + (0.05) 2 /(1) 2 } = 0.0025 E[ln(A)] = ln(E[A]) – 0.5 Var[ln(A)] = ln(1) – 0.5(0.0025) = -0.00125 Thus, E[ln(F)] = ln(40) – 0.00125 = 3.688 Var[ln(F)] = (0.1) 2 + 0.0025 = 0.0125 = (0.112) 2 Therefore, F ~ lognormal( E[ln(F)] = 3.688, Var[ln(F)] = (0.112) 2 ) b) What is the expected value of F? (looking for real space answer) E[F] = exp{E[ln(F)] + 0.5 Var[ln(F)]} = exp{3.688 + 0.5(0.112) 2 } = 40.2 c) What is P[F <= 35 kips]? P[F ≤ 35] = P[ln(F) ≤ ln(35)] =

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Hw__7_Solutions_to_POST - #1 Given E[ln(S = ln(40 ksi...

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