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**Unformatted text preview: **R ] – 0.5 Var[ln(R)] = ln[120] – 0.5 (0.0832) 2 = 4.78 Doing the same for ln[S] Var[ ln(S) ] = ln[ 1 + σ S 2 /μ S 2 ] = ln[ 1 + (15) 2 /75 2 ] = 0.0392 = (0.2) 2 E[ ln(S) ] = ln[μ S ] – 0.5 Var[ln(S)] = ln[75] – 0.5 (0.2) 2 = 4.30 Lecture 16; page 15 The variable of interest is U = ln(R) – ln(S) ~ N(μ lnR – μ lnS , σ lnR 2 + σ lnS 2 ) = N( 4.78 – 4.30, (0.0832) 2 + (0.2) 2 ) = N( 0.486, 0.046 ) = N( 0.486, (0.21) 2 ) P[ R/S < 1] = P[ ln(R) – ln(S) < 0 ] = P[ (U-μ U )/ σ U < –0.486/0.21 ] = P[ Z < –0.486/0.21 ] = P[ Z < – 2.26 ] = 0.012 = 1.2% (using tables) Using bivariate normal model obtained failure probability of 0.6% —> model choice matters....

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