Materi Lanjutan Distribusi Normal (Matematika Peminatan Kelas XII)

C. Transformasi Suatu Variabel Random Berdistribusi Normal

Dalam menentukan luas suatu variabel berdistribusi normal ke dalam variabel random berdistribusi normal baku dengan jalan mentransformasikannya

Adapun luasnya sama yaitu: 

$\begin{aligned}&\begin{aligned}\textrm{P}(\textrm{x}_{1}< X< \textrm{x}_{2})&=\displaystyle \int_{\textrm{x}_{1}}^{\textrm{x}_{2}}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}\left (\displaystyle \frac{\textrm{x}-\mu }{\sigma }  \right )^{2}}d\textrm{x}\\ &=\displaystyle \int_{\textrm{z}_{1}}^{\textrm{z}_{2}}\displaystyle \frac{1}{ \sqrt{2\pi }}e^{-\frac{1}{2}\textrm{Z}^{2}}d\textrm{x}\\ &=\textrm{P}(\textrm{z}_{1}<\textrm{Z}<\textrm{z}_{2}) \end{aligned}\\\\ &\textrm{Luas di atas adalah hasil tranformasi}\\ &\textrm{variabel acak X}\sim \textrm{N}(\mu ,\sigma )\: \: \textrm{ke}\: \: \textrm{Z}\sim \textrm{N}(0,1)\\ &\textrm{dengan}\: \: \textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma } \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui variabel acak Z berdistribusi}\\ &\textrm{normal}\: \: \textrm{N}(0,1)\: \textrm{dan X berdistribusi}\\ &\textrm{normal N}(18,5).\: \textrm{Tentukanlah besar}\\ &\textrm{peluang berikut}\\ &\textrm{a}.\quad \textrm{P}(\textrm{Z}>0,68)\\ &\textrm{b}.\quad \textrm{P}(\textrm{X}<20)\\ &\textrm{c}.\quad \textrm{P}(0,36<\textrm{Z}<1,42)\\ &\textrm{d}.\quad \textrm{P}(17<\textrm{X}<18,5)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \textrm{P}(\textrm{Z}>0,68)&=0,5-\textrm{P}(0<\textrm{Z}<0,68)\\ &=0,5-0,2517=\color{blue}0,2483 \end{aligned}\\&\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}20, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =18 \\  \sigma  & =5  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}20\color{black}-18}{5}=0,4,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(\textrm{X}< 20)&=\textrm{P}(\textrm{Z}< 0,4)\\ &=0,5+\textrm{P}(0< \textrm{Z}< 0,4)\\ &=0,5+0,1554=\color{blue}0,6554 \end{aligned} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{P}(0,36<\textrm{Z}<1,42)&=\textrm{P}(0<\textrm{Z}<1,42)-\textrm{P}(0<\textrm{Z}<0,36)\\ &=0,4222-0,1406=\color{blue}0,2816 \end{aligned}\\&\begin{aligned}\textrm{d}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}17, \: \: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}18,5\\ &\begin{cases} \mu  & =18 \\  \sigma  & =5  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}17\color{black}-18}{5}=-0,2,\: \: \textrm{dan}\\ &\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}18,5\color{black}-18}{5}=0,1,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(17<\textrm{X}< 18,5)&=\textrm{P}(-0,2<\textrm{Z}< 0,1)\\ &=\textrm{P}(0< \textrm{Z}< 0,2)+\textrm{P}(0< \textrm{Z}< 0,1)\\ &=0,0793+0,0398=\color{blue}0,1191 \end{aligned} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui variabel acak X berdistribusi}\\ &\textrm{normal memiliki rata-rata 16 dan simpangan}\\ &\textrm{baku}\: \: 1,4.\: \: \textrm{Hitunglah besar peluang dari}\\ &\textrm{a}.\quad \textrm{P}(\textrm{X}\leq 18,8)\\ &\textrm{d}.\quad \textrm{P}(12,1\leq \textrm{X}\leq 16,3)\\\\ &\textbf{Jawab}:\\&\begin{aligned}\textrm{a}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}18,8, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =16 \\  \sigma  & =1,4  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}18,8\color{black}-16}{1,4}=2,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(\textrm{X}< 18,8)&=\textrm{P}(\textrm{Z}< 2)\\ &=0,5+\textrm{P}(0< \textrm{Z}< 2)\\ &=0,5+0,4772=\color{blue}0,9772 \end{aligned} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}12,1 \: ,\: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}16,3\\ &\begin{cases} \mu  & =16 \\  \sigma  & =1,4  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}12,1\color{black}-16}{1,4}=-2,79,\: \: \textrm{dan}\\&\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}16,3\color{black}-16}{1,4}=0,21,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(12,1<\textrm{X}< 16,3)&=\textrm{P}(-2,79<\textrm{Z}< 0,21)\\ &=\textrm{P}(0< \textrm{Z}< 2,79)+\textrm{P}(0< \textrm{Z}< 0,21)\\ &=0,4974+0,0832=\color{blue}0,5806 \end{aligned} \end{aligned}   \end{array}$ 

$\begin{array}{ll}\\ 3.&\textrm{Di sebuah MA dengan 1000 siswa diperoleh data}\\ &\textrm{rata-rata berat bada siswanya}\: \: 54\: \textrm{Kg}\: \textrm{dan simpangan}\\ &\textrm{baku}\: 8\: \textrm{Kg}.\: \textrm{Jika data tersebut berdistribusi normal}\\ &\textrm{tentukan bsnysk siswa yang memiliki berat badan}\\ &\textrm{a}.\quad \textrm{lebih dari 70 Kg}\\ &\textrm{b}.\quad \textrm{antara 40 Kg sampai 50 Kg}\\\\ &\textbf{Jawab}:\\  &\begin{aligned}\textrm{a}.\quad&\textrm{Transformasi}\: \: \textrm{x}=\color{red}70, \: \: \color{black}\textrm{dengan}\begin{cases} \mu  & =54 \\  \sigma  & =8  \end{cases}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}70\color{black}-54}{8}=\frac{16}{8}=2\\ &\textrm{Dari tabel diperoleh}\\ &P(0< \textrm{Z}< 2)=0,4772,\: \: \textrm{selanjutnya}\\ &\begin{aligned}\textrm{P}(\textrm{X}>70)&=\textrm{P}(\textrm{Z}>2)\\ &=0,5-\textrm{P}(0< \textrm{Z}< 2)\\ &=0,5-0,4772=\color{blue}0,0228 \end{aligned}\\ &\textrm{Selanjutnya frekuensi harapan dalam hal ini}\\ &\begin{aligned}f_{h}(\textrm{X}>70)&=1000\times P(\textrm{X}>70)\\ &=1000\times \color{blue}0,0228\\ &=22,8\\ &\approx 23  \end{aligned}\\ &\textrm{Jadi, ada sebanyak 23 anak dengan berat badan}\\ &>\: \:  \textrm{70 Kg} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\textrm{Transformasi}\: \: \textrm{x}_{1}=\color{red}40 \: ,\: \color{black}\textrm{dan}\: \: \textrm{x}_{2}=\color{red}50\\ &\begin{cases} \mu  & =54 \\  \sigma  & =8  \end{cases}\\ &\textrm{z}_{1}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}40\color{black}-54}{8}=-1,75,\: \: \textrm{dan}\\&\textrm{z}_{2}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\displaystyle \frac{\color{red}50\color{black}-54}{8}=-0,5,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{P}(40<\textrm{X}<50)&=\textrm{P}(-1,75<\textrm{Z}<-0,5)\\ &=\textrm{P}(0,5<\textrm{Z}<1,75)\\ &=\textrm{P}(0< \textrm{Z}< 1,75)-\textrm{P}(0< \textrm{Z}< 0,5)\\ &=0,4599-0,1915=\color{blue}0,2684 \end{aligned}\\ &\textrm{Selanjutnya frekuensi harapan dalam hal ini}\\ &\begin{aligned}f_{h}(40<\textrm{X}<50)&=1000\times P(40<\textrm{X}<50)\\ &=1000\times \color{blue}0,2684\\&=268,4\\ &\approx 268  \end{aligned}\\ &\textrm{Jadi, ada sebanyak 268 anak dengan berat badan}\\ &\textrm{antara 40 samapi 50 Kg} \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui Nilai-nilai ujian penerimaan pegawai baru}\\&\textrm{diperoleh mean 78 dan deviasi standarnya 6. Jika}\\ &\textrm{hanya}\: \: 12,5\%\: \: \textrm{calon yang akan diterima, maka nilai}\\ &\textrm{terendah yang lolos jika distribusinya normal}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Transformasi}\: \: \textrm{X}=\color{red}\textrm{x}\\ &\color{black}\textrm{dengan}\begin{cases} \mu  & =78 \\  \sigma  & =6,\: \textrm{dan}\\ \textrm{P}(\textrm{X}>x)&=12,5\%=0,125  \end{cases}\\&\begin{aligned}\textrm{P}(\textrm{X}>\textrm{x})\qquad&=\textrm{P}(\textrm{Z}>0)-\textrm{P}(0< \textrm{Z}<\textrm{z})\\ 0,125&=0,5-\textrm{P}(0< \textrm{Z}< \textrm{z})\\ \textrm{P}(0< \textrm{Z}< \textrm{z})\: \, &=0,5-0,125=\color{blue}0,375 \end{aligned}\\ &\textrm{Dari tabel diperoleh}\: \: \textrm{P}(0< \textrm{Z}< \textrm{z})=0,375,\\ &\textrm{selanjutnya didapatkan nilai}\: \: z=\color{red}1,15\\ &\textrm{Sehingga nilai terendah x yang diterima}\\&\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }\\ &\Leftrightarrow \textrm{x}=\textrm{z}\sigma +\mu =(1,15).6+78=84,9\approx 85 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.