Diketahui \(F(18,25) = \Phi \left( {\frac{{\ln 18,25 – \mu }}{\sigma }} \right) = 0,2\)
\(F(35,8) = \Phi \left( {\frac{{\ln 35,8 – \mu }}{\sigma }} \right) = 0,8\) Rumus Standard Normal Distribution,
\(\Pr \left( {Z < z} \right) = \Phi \left( {\frac{{\ln z – \mu }}{\sigma }} \right)\) Step 1 Sumber: six-sigma-material.com Pada tabel distribusi normal diatas, 0,8 berada diantara z=0,84 dan z=0,85. Dengan rumus interpolasi,
\((0,8 – 0,7995)(0,05 – 0,04) = (0,8023 – 0,8)x\)
\(x = 0,002173913\)
\(x \cong 0,002\)
Maka 0,8 berada tepat saat z=0,842
Step 2 \(F(35,8) = \Phi \left( {\frac{{\ln 35,8 – \mu }}{\sigma }} \right) = 0,8\)
\(\frac{{\ln 35,8 – \mu }}{\sigma } = 0,842\) \(F(18,25) = \Phi \left( {\frac{{\ln 18,25 – \mu }}{\sigma }} \right) = 0,2\)
\(\frac{{\ln 18,25 – \mu }}{\sigma } = – 0,842\) \(\frac{{\frac{{\ln 35,8 – \mu }}{\sigma }}}{{\frac{{\ln 18,25 – \mu }}{\sigma }}} = \frac{{0,842}}{{ – 0,842}}\)
\(\ln 35,8 – \mu = – 1\left( {\ln 18,25 – \mu } \right)\)
\(\mu = 3,241056487\)
\(\mu \cong 3,241\) \(\frac{{\ln 35,8 – 3,241}}{\sigma } = 0,842\)
\(\sigma = 0,4001756454\)
\(\sigma \cong 0,4\) Maka \(\Pr \left( {Z > 30} \right) = 1 – \Pr (Z < 30)\)
\(\Pr \left( {Z > 30} \right) = 1 – \Phi \left( {\frac{{\ln 30 – 3,241}}{{0,4}}} \right)\)
\(\Pr \left( {Z > 30} \right) = 1 – \Phi \left( {0,4} \right)\)
\(\Pr \left( {Z > 30} \right) = 1 – 0,6554\)
\(\Pr \left( {Z > 30} \right) = 0,3446\) Jawaban d. 0,3446