Contoh Soal 3 Distribusi Normal

 $\begin{array}{l}\\ 11.& \text{Tersiar kabar bahwa harga beras di pasar di}\\ & \text{wilayah B adalah}\text{Rp}8.000,00/\text{Kg dengan}\\ & \text{simpangan baku}\text{Rp}1.500,00.\text{Berdasar kabar}\\ & \text{tersebut dilakukan penelitian dengan mengambil}\\ & \text{sampel secara acak sebanyak}60\text{kios yang dan}\\ & \text{diperoleh rata-rata harga beras}\text{Rp}7.800,00/\text{Kg}\\ & \text{Jika penghitungan menggunakan tingkat}\\ & \text{signifikansi}5\mathrm{\%},\text{maka kesimpulan berikut }\\ & \text{yang tepat adalah}....\\ & \text{a}.\text{harga beras di pasar lebih dari}\text{Rp}7.800,00/\text{Kg}\\ & \text{b}.\text{harga beras di pasar lebih dari}\text{Rp}8.000,00/\text{Kg}\\ & \text{c}.\text{harga beras di pasar kurang dari}\text{Rp}0.000,00/\text{Kg}\\ & \text{d}.\text{harga beras di pasar}\text{Rp}7.800,00/\text{Kg}\\ & \text{e}.\text{harga beras di pasar}\text{Rp}8.000,00/\text{Kg}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{―}{\text{Hipotesis}}\\ & \text{Rata-rata harga beras dipasar}\text{Rp}8.000,00/\text{Kg}\\ & {\text{H}}_0:\mu =8.000\\ & {\text{H}}_0:\mu \ne 8.000\\ & \underset{―}{\text{Daerah Kritis}}\\ & \text{Taraf nyata yang dipilih adalah}=\alpha =0,05=5\mathrm{\%}\\ & {\displaystyle \frac{\alpha }{2}=2,5\mathrm{\%}=0,025}\\ & {\text{z}}_{0,025}=1,96\\ & \text{maka daerah kritis/penolakannya adalah}\\ & \text{z}<-1,96\text{atau}\text{z}>1,96\end{array}\end{array}$.

$.\begin{array}{rl}& \underset{―}{\text{Nilai Satistik Uji}}\\ & \text{Dihitung dengan rumus}:\text{z}={\displaystyle \frac{\bar{\text{x}}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}}}\\ & \text{z}={\displaystyle \frac{7.800-8.000}{{\displaystyle \frac{1.500}{\sqrt{60}}}}=-{\displaystyle \frac{200\sqrt{60}}{1.500}=-1,03}}\\ & \underset{―}{\text{Keputusan Uji}}\\ & \text{Karena nilai}-1,96<\text{z}<1,96,\\ & \text{maka}{\text{H}}_0\text{diterima}\\ & \underset{―}{\text{Kesimpulan}}\\ & \text{Rata-rata harga beras dipasar}\text{Rp}8.000,00/\text{Kg}\end{array}$ .

Contoh Soal 2 Distribusi Normal

$\begin{array}{l}\\ 6.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,4861\\ & \text{b}.0,4878\\ & \text{c}.0,4881\\ & \text{d}.0,4938\\ & \text{e}.0,4946\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & \downarrow & & & & \\ ⋮& & & & & & & & & & \\ 2,2& & & & & & 0,4878& & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=2,25\text{luasnya}=0,4878\end{array}$.

$\begin{array}{l}\\ 7.& \text{Luas daerah yang diarsir di bawah}\\ & \text{kurva normal baku berikut adalah}....\\ & \text{a}.0,1138\\ & \text{b}.0,3810\\ & \text{c}.0,3862\\ & \text{d}.0,4948\\ & \text{e}.0,5000\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& & & & & & & \downarrow & & \downarrow & \\ ⋮& & & & & & & & & & \\ 1,1& & & & & & & \downarrow & & 0,3810& \\ ⋮& & & & & & & & & & \\ 2,5& & & & & & & 0,4948& & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga nilai}\text{z}=1,18\text{luasnya}=0,3810\\ & \text{Dan nilai}\text{z}=2,56\text{luasnya}=0,4948\\ & \text{maka luas arsiran}\\ & =P(1,18<Z<2,56)\\ & =P(0<Z<2,56)-P(0<Z<1,18)\\ & =0,4948-0,3810\\ & =0,1138\end{array}$.

$\begin{array}{l}\\ 8.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah adalah 0,9332, maka nilai}\text{z}=....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.1,05\\ & \text{b}.1,06\\ & \text{c}.1,16\\ & \text{d}.1,50\\ & \text{e}.1,60\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran adalah}\\ & =P(Z<z)=0,9332\\ & =0,5+0,4332\\ & =0,5+P(0<Z<z)\\ & \text{lihat/konfirmasi ke tabel}\\ & \text{z}=1,50\\ & \begin{array}{rl}& \text{Perhatikan tabel}\text{distribusi normal}\text{berikut}\\ & \begin{array}{|ccccccccccc|}\hline \text{z}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ ⋮& \downarrow & & & & & & & & & \\ ⋮& & & & & & & & & & \\ 1,5& 0,4332& & & & & & & & & \\ ⋮& & & & & & & & & & \\ \hline\end{array}\\ & \text{Sehingga luasnya}=0,4878,\text{batas z}=1,50\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Pada suatu kelas seorang guru matematika}\\ & \text{menyatakan bahwa nilai ulangan mapel }\\ & \text{yang diampunya tidak kurang dari}68\\ & \text{Untuk menentukan uji tersebut, maka guru}\\ & \text{yang bersangkutan memilih 10 anak secara}\\ & \text{acak untuk ditanyai nilai hasil ulangannya}\\ & \text{Hipotesis}{\text{H}}_0\text{dan}{\text{H}}_1\text{yang tepat dari kondisi}\\ & \text{kondisi di atas adalah}....\\ & \text{a}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ & \text{b}.\begin{array}{c}{\text{H}}_0:\mu =68\\ {\text{H}}_1:\mu >68\end{array}\\ & \text{c}.\begin{array}{c}{\text{H}}_0:\mu \ge 68\\ {\text{H}}_1:\mu <68\end{array}\\ & \text{d}.\begin{array}{c}{\text{H}}_0:\mu \le 68\\ {\text{H}}_1:\mu >q68\end{array}\\ & \text{e}.\begin{array}{c}{\text{H}}_0:\mu >68\\ {\text{H}}_1:\mu \ne 68\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \text{Cukup jelas}\\ & \text{Dan ini contoh uji satu pihak, yaitu kiri}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Seorang petugas}customerservice\text{menyatakan}\\ & \text{bahwa di dealer A dapat mensevis rata-rata }\\ & \text{sebanyak}75\text{unit sepeda motor perhari}.\text{Untuk}\\ & \text{mengecek kebenaran pernyataan di atas diambil}\\ & \text{sampel beberapa hari secara random sebanyak }\\ & \text{20 hari. Dari penelitian ini diperoleh rata-rata}\\ & \text{78 unit dengan simpangan bakunya 5. Hasil}\\ & \text{perhitungan}{\text{z}}_0-\text{nya adalah}....\\ & \text{a}.2,35\\ & \text{b}.2,43\\ & \text{c}.2,55\\ & \text{d}.2,68\\ & \text{e}.2,75\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui data dianggap berdistribusi}\\ & \text{normal baku}N(0,1)\text{dengan}\\ & \text{Rata-rata sampel}=\bar{\text{x}}=78\text{unit sepeda motor}\\ & \text{Rata-rata populasinya yang diuji}={\mu }_0=75\\ & \text{Simpangan bakunya}=\sigma =5\text{unit}\\ & \text{dengan banyak data sampel}=n=20\text{hari}\\ & \text{Penghitungan nilainya}z-\text{nya}\\ & ={\displaystyle \frac{\bar{\text{x}}-{\mu }_0}{{\displaystyle \frac{\sigma }{\sqrt{n}}}}}\\ & ={\displaystyle \frac{78-75}{{\displaystyle \frac{5}{\sqrt{20}}}}={\displaystyle \frac{3}{{\displaystyle \frac{5}{2\sqrt{5}}}}}}\\ & ={\displaystyle \frac{6}{\sqrt{5}}=\frac{6}{2,236}=2,683}\end{array}\end{array}$.

Contoh Soal 1 Distribusi Normal

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

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

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

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

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

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: 

$\begin{array}{rl}& \begin{array}{rl}\text{P}({\text{x}}_1<X<{\text{x}}_2)& ={\displaystyle {\int }_{{\text{x}}_1}^{{\text{x}}_2}{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}d\text{x}}}\\ & ={\displaystyle {\int }_{{\text{z}}_1}^{{\text{z}}_2}{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{\text{Z}}^2}d\text{x}}}\\ & =\text{P}({\text{z}}_1<\text{Z}<{\text{z}}_2)\end{array}\\ \\ & \text{Luas di atas adalah hasil tranformasi}\\ & \text{variabel acak X}\sim \text{N}(\mu ,\sigma )\text{ke}\text{Z}\sim \text{N}(0,1)\\ & \text{dengan}\text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui variabel acak Z berdistribusi}\\ & \text{normal}\text{N}(0,1)\text{dan X berdistribusi}\\ & \text{normal N}(18,5).\text{Tentukanlah besar}\\ & \text{peluang berikut}\\ & \text{a}.\text{P}(\text{Z}>0,68)\\ & \text{b}.\text{P}(\text{X}<20)\\ & \text{c}.\text{P}(0,36<\text{Z}<1,42)\\ & \text{d}.\text{P}(17<\text{X}<18,5)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\text{P}(\text{Z}>0,68)& =0,5-\text{P}(0<\text{Z}<0,68)\\ & =0,5-0,2517=0,2483\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}\text{x}=20,\text{dengan}\{\begin{array}{ll}\mu & =18\\ \sigma & =5\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{20-18}{5}=0,4,\text{maka}}}\\ & \begin{array}{rl}\text{P}(\text{X}<20)& =\text{P}(\text{Z}<0,4)\\ & =0,5+\text{P}(0<\text{Z}<0,4)\\ & =0,5+0,1554=0,6554\end{array}\end{array}\\ & \begin{array}{rl}\text{c}.\text{P}(0,36<\text{Z}<1,42)& =\text{P}(0<\text{Z}<1,42)-\text{P}(0<\text{Z}<0,36)\\ & =0,4222-0,1406=0,2816\end{array}\\ & \begin{array}{rl}\text{d}.& \text{Transformasi}{\text{x}}_1=17,\text{dan}{\text{x}}_2=18,5\\ & \{\begin{array}{ll}\mu & =18\\ \sigma & =5\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{17-18}{5}=-0,2,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{18,5-18}{5}=0,1,\text{maka}}}\\ & \begin{array}{rl}\text{P}(17<\text{X}<18,5)& =\text{P}(-0,2<\text{Z}<0,1)\\ & =\text{P}(0<\text{Z}<0,2)+\text{P}(0<\text{Z}<0,1)\\ & =0,0793+0,0398=0,1191\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui variabel acak X berdistribusi}\\ & \text{normal memiliki rata-rata 16 dan simpangan}\\ & \text{baku}1,4.\text{Hitunglah besar peluang dari}\\ & \text{a}.\text{P}(\text{X}\le 18,8)\\ & \text{d}.\text{P}(12,1\le \text{X}\le 16,3)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Transformasi}\text{x}=18,8,\text{dengan}\{\begin{array}{ll}\mu & =16\\ \sigma & =1,4\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{18,8-16}{1,4}=2,\text{maka}}}\\ & \begin{array}{rl}\text{P}(\text{X}<18,8)& =\text{P}(\text{Z}<2)\\ & =0,5+\text{P}(0<\text{Z}<2)\\ & =0,5+0,4772=0,9772\end{array}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}{\text{x}}_1=12,1,\text{dan}{\text{x}}_2=16,3\\ & \{\begin{array}{ll}\mu & =16\\ \sigma & =1,4\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{12,1-16}{1,4}=-2,79,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{16,3-16}{1,4}=0,21,\text{maka}}}\\ & \begin{array}{rl}\text{P}(12,1<\text{X}<16,3)& =\text{P}(-2,79<\text{Z}<0,21)\\ & =\text{P}(0<\text{Z}<2,79)+\text{P}(0<\text{Z}<0,21)\\ & =0,4974+0,0832=0,5806\end{array}\end{array}\end{array}$ 

$\begin{array}{l}\\ 3.& \text{Di sebuah MA dengan 1000 siswa diperoleh data}\\ & \text{rata-rata berat bada siswanya}54\text{Kg}\text{dan simpangan}\\ & \text{baku}8\text{Kg}.\text{Jika data tersebut berdistribusi normal}\\ & \text{tentukan bsnysk siswa yang memiliki berat badan}\\ & \text{a}.\text{lebih dari 70 Kg}\\ & \text{b}.\text{antara 40 Kg sampai 50 Kg}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Transformasi}\text{x}=70,\text{dengan}\{\begin{array}{ll}\mu & =54\\ \sigma & =8\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{70-54}{8}=\frac{16}{8}=2}}\\ & \text{Dari tabel diperoleh}\\ & P(0<\text{Z}<2)=0,4772,\text{selanjutnya}\\ & \begin{array}{rl}\text{P}(\text{X}>70)& =\text{P}(\text{Z}>2)\\ & =0,5-\text{P}(0<\text{Z}<2)\\ & =0,5-0,4772=0,0228\end{array}\\ & \text{Selanjutnya frekuensi harapan dalam hal ini}\\ & \begin{array}{rl}{f}_h(\text{X}>70)& =1000\times P(\text{X}>70)\\ & =1000\times 0,0228\\ & =22,8\\ & \approx 23\end{array}\\ & \text{Jadi, ada sebanyak 23 anak dengan berat badan}\\ & >\text{70 Kg}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}{\text{x}}_1=40,\text{dan}{\text{x}}_2=50\\ & \{\begin{array}{ll}\mu & =54\\ \sigma & =8\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{40-54}{8}=-1,75,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{50-54}{8}=-0,5,\text{maka}}}\\ & \begin{array}{rl}\text{P}(40<\text{X}<50)& =\text{P}(-1,75<\text{Z}<-0,5)\\ & =\text{P}(0,5<\text{Z}<1,75)\\ & =\text{P}(0<\text{Z}<1,75)-\text{P}(0<\text{Z}<0,5)\\ & =0,4599-0,1915=0,2684\end{array}\\ & \text{Selanjutnya frekuensi harapan dalam hal ini}\\ & \begin{array}{rl}{f}_h(40<\text{X}<50)& =1000\times P(40<\text{X}<50)\\ & =1000\times 0,2684\\ & =268,4\\ & \approx 268\end{array}\\ & \text{Jadi, ada sebanyak 268 anak dengan berat badan}\\ & \text{antara 40 samapi 50 Kg}\end{array}\end{array}$.

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

DAFTAR PUSTAKA

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

Distribusi Normal (Matematika Peminatan Kelas XII)

 A. Pengertian Distribusi Normal

Distribusi normal adalah salah satu distribusi model variabel acak kontinue yang sangat penting dalam probabilitas.

Distribusi normal yang juga dikenal dengan distribusi Gaussian ini memiliki grafik berbentuk bel/lonceng yang selanjutnya juga dikenal dengan kurva normal karena bentuk kurvanya seperti lonceng. Persamaan kurva  tersebut dinamakan dengan fungsi distribusi normal. Adapun fungsi distribusi normal untuk variabel acak kontinue X didefinisikan dengan.

$\begin{array}{rl}& f(x)={\displaystyle \frac{1}{\sigma \sqrt{2\pi }}.e^{{.}^{-\frac{1}{2}{\left({\displaystyle \frac{x-\mu }{\sigma }}\right)}^2}}}\\ & \text{Dengan}\\ & \sigma :\text{parameter untuk standar deviasi}\\ & \mu :\text{parameter untuk rata-rata (mean)}\\ & e:\text{Kontanta alam (2,718...)}\\ & \text{Dengan domain fungsi}f-\mathrm{\infty }<x<\mathrm{\infty }\end{array}$.

Berikut gambar kurva normalnya $\text{N}(0,1)$

Untuk variabel acak X berdistribusi normal dilambangkan dengan $\text{X}\sim \text{N}(\mu ,\sigma )$. Selanjutnya jika $\mu =0$ dan $\sigma =1$, maka akan diperoleh distribusi normal baku (standar) yaitu $\text{N}(0,1)$. Dan rumus fungsi variabel acak Z yang berdistribusi normal  baku adalahh: $f(z)={\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}}$.

Karena kurva di atas adalah kurva dari grafik fungsi peluang, maka luas yang dibatasi adalah garfik fungsi dan sumbu mendatarnya adalah berharga 1, atau dapat juga dituliskan

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(z)dz={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}dz=1}$.

Karena grafik simetris terhadap garis $\mu =0$, maka luas di kiri dan kanan garis $\mu =0$ bernilai $0,5$ atau

${\int }_{-\mathrm{\infty }}^{0}f(z)dz={\int }_{-\mathrm{\infty }}^0{\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}dz=0,5}$ dan ${\int }_0^{\mathrm{\infty }}f(z)dz={\int }_0^{\mathrm{\infty }}{\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}dz=0,5}$.

B. Penghitungan luas di Bawah Kurva Distribusi Normal

B. 1 Penghitungan luasan di bawah kurva

Penentuan luas wilayah ini sangatlah tidak mudah karena melibatkan banyak aspek, tetapi ada cara lain dalam penentuan luas daerah di bawah kurva normal, yaitu dengan bantuan tabel distribusi $\text{Z}$ sebagaimana tabel sederhana berikut

Sumber dari gambar di atas adalah dari screenshot dari youtube Channel Ari Susanti  

B. 2 Penghitungan luasan di bawah dengan Interval  Tertentu

Luasan daerah dibawah kurva normal baku pada interval  $z_1<\text{Z}<z_2$ dapat dituliskan sebagai  $P(z_1<\text{Z}<z_2)={\displaystyle {\int }_{z_1}^{z_2}{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{{.}^{-\frac{1}{2}{Z}^2}}dz}}$.

Perhatikanlah ilustrasi berikut ini

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Perhatikanlahdaerah berarsir pada kurva normal}\\ & \text{berikut untuk interval}0<\text{Z}<1,25\end{array}$.

$.\begin{array}{rl}& \text{a}.\text{Nyatakan dengan bentuk integral yang menyatakan}\\ & \text{luas daerah yang terarsir}\\ & \text{b}.\text{Tentukan luas daerah yang diarsir dengan bantuan}\\ & \text{tabel distribusi normal baku}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Diketahui fungsi normal baku dalam variabel}z\text{adalah}:\\ & f(z)={\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}}\\ & \text{maka daerah yang diarsir pada interval}0<\text{Z}<1,25\\ & \text{Yaitu}:\\ & L={\displaystyle {\int }_0^{1,25}f(z)dz={\displaystyle {\int }_0^{1,25}{\displaystyle \frac{1}{\sqrt{2\pi }}{\text{e}}^{{.}^{-\frac{1}{2}{Z}^2}}dz}}}\\ \text{b}.& \text{Adapaun cara tabel adalah sebagai berikut}\\ & \text{Lihat gambar di atas, yaitu}:0,3944\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Pada interval berikut, tentukanlah luas}\\ & \text{daerah dibawah kurva normbal baku}\\ & \text{a}.\text{Z}>0,96\\ & \text{b}.-0,72<\text{Z}<2,08\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Karena luas daerah di kanan garis}z=0\\ & \text{maka luas}:0,96<\text{Z}<\mathrm{\infty }\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}6\\ \downarrow \end{array}\\ 0& 0,9\to & 0,3315\\ \hline\end{array}\\ & \text{Jadi, luasnya}=0,5-0,3315=0,1685\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Karena luas daerah di kiri dan kanan garis}z=0\\ & \text{maka luas}:-0,72<\text{Z}<2,08\text{atau}\text{P}(-0,72<\text{Z}<2,08)\\ & \underset{―}{\text{Untuk}}:-0,72<\text{Z}<0=0<\text{Z}<0,72\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}2\\ \downarrow \end{array}\\ 0& 0,7\to & 0,2642\\ \hline\end{array}\\ & \underset{―}{\text{Sedangkan untuk}}:0<\text{Z}<2,08\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}8\\ \downarrow \end{array}\\ 0& 2,0\to & 0,4812\\ \hline\end{array}\\ & \text{Jadi, luasnya}=0,2642+0,4812=0,7454\\ & \text{Berikut ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah besar peluang dari variabel}\\ & \text{variabel acak Z berdistribusi normal baku}\\ & \text{a}.\text{P}(\text{Z}<1,2)\\ & \text{b}.\text{P}(0,32<\text{Z}<1,5)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{3. a. Perhatikan ilustrasi berikut ini}\end{array}$.

$.\begin{array}{rl}& \text{Karena luas daerah di kiri dan kanan garis}z=0\\ & \text{maka luas}:\text{P}(\text{Z}<1,2)=\text{P}(-\mathrm{\infty }<\text{Z}<1,2)\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}0\\ \downarrow \end{array}\\ 0& 1,2\to & 0,3849\\ \hline\end{array}\\ & \text{Jadi, luasnya}=0,5+0,3315=0,8849\end{array}$.

$.\begin{array}{rl}3.b& \text{Untuk}\text{P}(0,32<\text{Z}<1,5)\\ & \text{Perhatikan ilsutrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Karena luas daerah di kanan garis}z=0\\ & \text{maka luas}:0,32<\text{Z}<1,5\\ & \underset{―}{\text{Untuk}}:0<\text{Z}<0,32\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}2\\ \downarrow \end{array}\\ 0& 0,3\to & 0,1255\\ \hline\end{array}\\ & \underset{―}{\text{Sedangkan untuk}}:0<\text{Z}<1,5\\ & \begin{array}{|ccc|}\hline z& z& \begin{array}{c}0\\ \downarrow \end{array}\\ 0& 1,5\to & 0,4332\\ \hline\end{array}\\ & \text{Jadi, luasnya}=0,4332-0,1255=0,3077\end{array}$ .

DAFTAR PUSTAKA

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