Diketahui \(\mu = 80\), dan \(\sigma=4\text{,}8\)
Pertanyaan diatas dapat diselesaikan dengan tabel Z.
Dengan menggunakan rumus
\[\begin{aligned}Z&=\frac{x-\mu}{\sigma }\end{aligned}\]
a. \[\begin{aligned}P(X\leq87\text{,}2)&=P(Z\leq87\text{,}2)\\&=P\left(Z\leq\frac{87\text{,}2-80}{4\text{,}8}\right)\\&=P\left(Z\leq1\text{,}5\right)\\&=0\text{,}9332\end{aligned}\]
b.\[\begin{aligned}P(X\geq76\text{,}4)&=1-P(Z\leq76\text{,}4)\\&=1-P\left(Z\leq\frac{76\text{,}4-80}{4\text{,}8}\right)\\&=1-P\left(Z\leq-0\text{,}75\right)\\&=1-0\text{,}2266\\&=0\text{,}774\end{aligned}\]
c.\[\begin{aligned}P(81\text{,}2<X<86)&=P(Z<86)-P(Z<81\text{,}2)\\&=P(Z<\frac{86-80}{4\text{,}8})P\left(Z<\frac{81\text{,}2-80}{4\text{,}8}\right)\\&=P\left(Z<1\text{,}25\right)-P\left(Z<0\text{,}25\right)\\&=0\text{,}8944-0\text{,}5987\\&=0\text{,}2957\end{aligned}\]
d.\[\begin{aligned}P(71\text{,}6<X<88\text{,}4)&=P(Z<88\text{,}4)-P(Z<71\text{,}6)\\&=P(Z<\frac{88\text{,}4-80}{4\text{,}8})P\left(Z<\frac{71\text{,}6-80}{4\text{,}8}\right)\\&=P\left(Z<1\text{,}75\right)-P\left(Z<-1\text{,}75\right)\\&=0\text{,}9599-0\text{,}0401\\&=0\text{,}9198\end{aligned}\]
e. Nilai terendah dari 12,5% nilai tertinggi = x
\(P(Z>z)\) =\(12\text{,}5\)% (dari tabel Z diperoleh z=1,54) sehingga x dihitung dengan rumus
\[\begin{aligned}Z&=\frac{{x}-\mu}{\sigma }\\1\text{,}54&=\frac{x-80}{4\text{,}8}\\7\text{,}4&=x-80\\x&=87\text{,}4\end{aligned}\]
Maka nilai terendah 12,5% nilai tertinggi adalah \(87\text{,}4\)