Pembahasan Ujian PAI: A70 – No. 17 – Mei 2018

\(L = f(11)f(15,2)f(18)f(21)f(25,8)\)

\(L = \left[ {{e^{ – \frac{1}{{2(11)}}{{(11 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(15,2)}}{{(15,2 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(18)}}{{(18 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(21)}}{{(21 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(25,8)}}{{(25,8 – \mu )}^2}}}} \right]\)

\(L = \left[ {{e^{ – \frac{1}{{2(11)}}{{(11 – \mu )}^2} – \frac{1}{{2(15,2)}}{{(15,2 – \mu )}^2} – \frac{1}{{2(18)}}{{(18 – \mu )}^2} – \frac{1}{{2(21)}}{{(21 – \mu )}^2} – \frac{1}{{2(25,8)}}{{(25,8 – \mu )}^2}}}} \right]\)

\(\frac{1}{{2(21)}}{(21 – \mu )^2} – \frac{1}{{2(25,8)}}{(25,8 – \mu )^2}\)

\(\frac{{d\ln (L)}}{{d\mu }} = \frac{{(11 – \mu )}}{{(11)}} + \frac{{(15,2 – \mu )}}{{(15,2)}} + \frac{{(18 – \mu )}}{{(18)}} + \frac{{(21 – \mu )}}{{(21)}} + \frac{{(25,8 – \mu )}}{{(25,8)}}\)

• \(\frac{{d\ln (L)}}{{d\mu }} = 0\)

\(0 = 1 – \frac{1}{{11}}\mu + 1 – \frac{1}{{15,2}}\mu + 1 – \frac{1}{{18}}\mu + 1 – \frac{1}{{21}}\mu + 1 – \frac{1}{{25,8}}\mu \)
\(0 = 5 – 0,29863\mu \)
\(\hat \mu = 16,74312695\)
\(\hat \mu = 16,74\)