Distribusi normal variabel random

Question

Suatu variable random mempunyai distribusi  normal dengan Mean 80, dan simpangan baku  4,8. Berapa probabilitasnya bahwa variable random tersebut akan mempunyai nilai :
a. Maksimal 87,2
b. Minimal  76,4
c. Antara 81,2 s/d 86
d. Antara 71,6 dan 88,4
e. berapa nilai terendah dari 12,5 % nilai tertinggi

rahmayetty · Answer

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 $87\text{,}4$