Contoh Soal Distribusi Normal

 $\begin{array}{ll}\\ 1.&\textrm{Fungsi distribusi normal variabel acak X}\\ &\textrm{dengan}\: \: \mu =8\: \: \textrm{dan}\: \: \sigma =2\: \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{2}}}\\ &\textrm{b}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{4}}}\\&\textrm{c}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{2}}}\\&\textrm{d}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\frac{(x-8)^{2}}{4}}}\\&\textrm{e}.\quad \color{red}\displaystyle f(x)=\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\frac{(x-8)^{2}}{8}}}\\\\&\textbf{Jawab}:\quad \textbf{e}\\&\begin{aligned}\displaystyle f(x)&=\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left (\frac{x-\mu}{\sigma } \right )^{2}}},\: \:  \textrm{dengan}\: \: \left\{\begin{matrix} \mu =8\\ \sigma =2 \end{matrix}\right.\\&=\displaystyle \frac{1}{2 \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left (\frac{x-8}{2 } \right )^{2}}}\\ &=\color{red}\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\displaystyle \frac{(x-8)^{2}}{8}}} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika variabel acak}\: \: Z\: \: \textrm{berdistribusi normal}\\ &\textrm{N}(0,1),\:  \textrm{nilai}\: \: \textrm{P}(Z< 2)\: \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{b}.\quad \displaystyle \int_{2}^{\infty }\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{c}.\quad \color{red}\displaystyle \int_{-\infty }^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{d}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left ( \displaystyle \frac{\textrm{x}-\mu }{\sigma } \right )^{2}}}\: dz\\ &\textrm{e}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left ( \displaystyle \frac{\textrm{x}-\mu }{\sigma } \right )^{2}}}\: dz\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&P(\textrm{Z}<2)\: ,\qquad \textrm{Z}\sim \textrm{N}(0,1)\\ &=P(-\infty <\textrm{Z}<0)+P(0<\textrm{Z}<2)\\ &=\color{red}\displaystyle \int_{-\infty }^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz \end{aligned}  \end{array}$.

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

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

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