Pembahasan Ujian PAI: A70 – No. 12 – Mei 2018

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

• \(F(35,8) = \Phi \left( {\frac{{\ln 35,8 – \mu }}{\sigma }} \right) = 0,8\)
  \(\frac{{\ln 35,8 – \mu }}{\sigma } = 0,842\)
• \(F(18,25) = \Phi \left( {\frac{{\ln 18,25 – \mu }}{\sigma }} \right) = 0,2\)
  \(\frac{{\ln 18,25 – \mu }}{\sigma } = – 0,842\)

• \(\frac{{\frac{{\ln 35,8 – \mu }}{\sigma }}}{{\frac{{\ln 18,25 – \mu }}{\sigma }}} = \frac{{0,842}}{{ – 0,842}}\)

  \(\ln 35,8 – \mu = – 1\left( {\ln 18,25 – \mu } \right)\)
  \(\mu = 3,241056487\)
  \(\mu \cong 3,241\)