Contoh Soal Distribusi Normal

 $\begin{array}{l}\\ 1.& \text{Fungsi distribusi normal variabel acak X}\\ & \text{dengan}\mu =8\text{dan}\sigma =2\text{adalah}....\\ & \text{a}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{2}}}}}\\ & \text{b}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{4}}}}}\\ & \text{c}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{2}}}}}\\ & \text{d}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{4}}}}}\\ & \text{e}.{\displaystyle f(x)={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-\frac{(x-8{)}^2}{8}}}}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle f(x)}& ={\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}}},\text{dengan}\{\begin{array}{c}\mu =8\\ \sigma =2\end{array}}\\ & ={\displaystyle \frac{1}{2\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left(\frac{x-8}{2}\right)}^2}}}}\\ & ={\displaystyle \frac{1}{\sqrt{8\pi }}{e}^{{.}^{-{\displaystyle \frac{(x-8{)}^2}{8}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika variabel acak}Z\text{berdistribusi normal}\\ & \text{N}(0,1),\text{nilai}\text{P}(Z<2)\text{adalah}....\\ & \text{a}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{b}.{\displaystyle {\int }_2^{\mathrm{\infty }}{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{c}.{\displaystyle {\int }_{-\mathrm{\infty }}^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\\ & \text{d}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}}}dz}}\\ & \text{e}.{\displaystyle {\int }_0^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}}}dz}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& P(\text{Z}<2),\text{Z}\sim \text{N}(0,1)\\ & =P(-\mathrm{\infty }<\text{Z}<0)+P(0<\text{Z}<2)\\ & ={\displaystyle {\int }_{-\mathrm{\infty }}^2{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-{\displaystyle \frac{1}{2}{\displaystyle z^2}}}}dz}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika luas daerah di bawah kurva}\\ & \text{berdistribusi normal pada interval}\text{Z}>z\\ & \text{adalah}L,\text{nilai}\text{P}(-z<Z<z)\text{adalah}....\\ & \text{a}.{\displaystyle 0,5+L}\\ & \text{b}.0,5-L\\ & \text{c}.{\displaystyle 1-L}\\ & \text{d}.{\displaystyle 1-2L}\\ & \text{e}.{\displaystyle 2L}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}P& (-z<Z<z)\\ & =0,5-L+0,5-L\\ & =1-2L\\ & \text{Berikut ilustrasi kurva beserta luasnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}\text{X}\sim \text{N}(20,4)\text{dan}Z\sim N(0,1)\\ & \text{Jika}P(0<Z<1)=0,3413,\text{maka nilai}\\ & P(X<24)\text{adalah}....\\ & \text{a}.{\displaystyle 0,1587}\\ & \text{b}.{\displaystyle 0,3174}\\ & \text{c}.{\displaystyle 0,3413}\\ & \text{d}.{\displaystyle 0,6826}\\ & \text{e}.{\displaystyle 0,8413}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}X\sim N(20,4)\{\begin{array}{ll}\mu & =20\\ \sigma & =4\end{array}\\ & \text{Dan diketahui pula}P(0<Z<1)=0,3413\\ & \text{Jika}Z\sim N(0,1),\text{maka untuk}P(X<24)\\ & \text{transformasi}\text{x}=24\text{menjadi}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }=\frac{24-20}{4}=\frac{4}{4}=1}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}P(X<24)& =P(Z<1)\\ & =0,5+P(0<Z<1)\\ & =0,5+0,3413\\ & =0,8413\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai kuartil atas dari data}\\ & \text{berdistribusi normal baku}=q\\ & \text{Pernyataan berikut yang tepat adalah}....\\ & \text{a}.\text{Luas daerah pada}(Z<q)=0,25\\ & \text{b}.\text{Luas daerah pada}(Z>q)=0,25\\ & \text{c}.\text{Luas daerah pada}(0<Z<q)=0,25\\ & \text{d}.\text{Luas daerah pada}(Z<-0,25)=q\\ & \text{e}.\text{Luas daerah pada}(0<Z<0,25)=q\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \text{Pembahasan diserahkan kepada pembaca}\\ & \text{yang budiman}\end{array}$.