Nilai Mean dan Proporsi

Question

Suatu sampel random sebanyak 40 ruang kelas diambil dari Kampus A STEI yang mempunyai luas rata-rata populasi atau mean (μ) 36,4 meter persegi (m2) dan simpangan baku atau standard deviation (σ) 5,2 m2. Hitunglah probabilitas luas rata-rata ruang kelas sampel tersebut :
(1) Antara 36 m2 dan 38 m2.
(2) 36,4 m2 sampai 40 m2.
(3) Paling luas 37,5 m2.
(4) Dari 37 m2 hingga 40 m2.
(5) Paling-paling 36 m2.

rahmayetty · Answer

Diketahui $\mu = 36\text{,}4$  , dan  $\sigma = 5\text{,}2$
Pertanyaan diatas dapat diselesaikan menggunakan Tabel Z
 $$\begin{aligned}Z&=\frac{X-\mu}{\sigma }\end{aligned}$$
1. $$\begin{aligned}P(36<X<38)&=P(X<38)-P(X<36)\&=P\left(Z<\frac{38-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{36-36\text{,}4}{5\text{,}2}\right)\&=P(Z<0\text{,}31)-P(Z<-0\text{,}07)\&=0\text{,}6217-0\text{,}4721\&=0\text{,}1496\end{aligned}$$
2.$$\begin{aligned}P(36\text{,}4<X<40)&=P(X<40)-P(X<36\text{,}4)\&=P\left(Z<\frac{40-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{36\text{,}4-36\text{,}4}{5\text{,}2}\right)\&=P(Z<0\text{,}69)-P(Z<0)\&=0\text{,}7549-0\text{,}5\&=0\text{,}2549\end{aligned}$$
3.$$\begin{aligned}P(X<37\text{,}5)&=P\left(Z<\frac{37\text{,}5-36\text{,}4}{5\text{,}2}\right)\&=P(Z<0\text{,}21)\&=0\text{,}5832\end{aligned}$$
4. $$\begin{aligned}P(37<X<40)&=P(X<40)-P(X<37)\&=P\left(Z<\frac{40-36\text{,}4}{5\text{,}2}\right)-P\left(Z<\frac{37-36\text{,}4}{5\text{,}2}\right)\&=P(Z<0\text{,}69)-P(Z<0\text{,}12)\&=0\text{,}7549-0\text{,}5478\&=0\text{,}2071\end{aligned}$$
5.$$\begin{aligned}P(X<36)&=P\left(Z<\frac{36-36\text{,}4}{5\text{,}2}\right)\&=P(Z<-0\text{,}07)\&=0\text{,}4721\end{aligned}$$