Distribusi Normal

A. Fungsi Probabilitas Kontinu

Jika pada distribusi peluang diskrit nilai  x diperjelas lagi menjadi nilai eksak atau kontinue, maka distribusi peluangnya akan berubah menjadi distribusi peluang kontinu.

Luas seluruh daerah di dalam kurva memiliki luas 1. Luas daerah pada wilayah yang diarsi (warna kuning) yang terletak antara X=a  dan X=b dapat dinyatakan dengan :  $P(a\le X\le b)={\displaystyle {\int }_a^{b}f(x)dx}$.

Sehingga peluang untu semua nilai x yang berada pada selang  $(a,b)$ adalah sama dengan luas kerapatan di bawah kurva antara batas  $x=a$  dan  $x=b$.

• $0\le f(x)\le 1$ untuk setiap nilai $x$.
• ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\displaystyle f(x)dx=1}$
• $P(a\le x\le b)={\int }_a^b{\displaystyle f(x)dx}$.

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Fungsi peluang lama bicara seorang}\\ & \text{operator sebagai berikut}\\ & f(x)=\{\begin{array}{ll}kx& \text{untuk}0\le k\le 5\\ k(10-x)& \text{untuk}5\le k\le 10\\ 0& \text{untuk}x\text{yang lain}\end{array}\\ & \text{Tentukanlah}\\ & \text{a}.\text{Nilai}k\\ & \text{b}.\text{Peluang operator telpon berbicara}\\ & \text{lebih dari 8 menit}\\ & \text{Peluang operator telpon berbicara}\\ & \text{2 sampai 4 menit}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Karena}f(x)\text{adalah fungsi peluang, maka}\\ & {\displaystyle {\int }_0^{5}kxdx+{\int }_5^{10}k(10-x)dx=1}\\ & \iff {\left[{\displaystyle \frac{1}{2}k{x}^2}\right]}_0^5+{[10kx-{\displaystyle \frac{1}{2}k{x}^2}]}_5^{10}=1\\ & \iff {\displaystyle \frac{1}{2}k(5^2-0^2)+(10k(10-5)-{\displaystyle \frac{1}{2}k({10}^2-5^2)})=1}\\ & \iff {\displaystyle \frac{1}{2}k(25)+10k(5)-{\displaystyle \frac{1}{2}k(100-25)=1}}\\ & \iff {\displaystyle \frac{25}{2}k+50k-{\displaystyle \frac{75}{2}k=1}}\\ & \iff 50k-25k=1\\ & \iff 25k=1\\ & \iff k={\displaystyle \frac{1}{25}}\\ \text{b}.& \text{Misalkan saja}\\ & X=\text{lama operator telpon bicara}\\ & \text{Peluang operator berbicara lebih}\\ & \text{dari 8 menit}=P(X>8),\\ & P(X>8)=P(8<X\le 10)\\ & ={\displaystyle {\int }_8^{10}k(10-x)dx}\\ & ={\displaystyle {\int }_8^{10}\frac{1}{25}(10-x)dx}\\ & ={\displaystyle \frac{1}{25}{[10x-{\displaystyle \frac{1}{2}{x}^2}]}_8^{10}}\\ & ={\displaystyle \frac{1}{25}(10(10-8)-{\displaystyle \frac{1}{2}({10}^2-8^2)})}\\ & ={\displaystyle \frac{1}{25}(10.(2)-{\displaystyle \frac{1}{2}(100-64)})}\\ & ={\displaystyle \frac{1}{25}(20-{\displaystyle \frac{1}{2}(36)})}\\ & ={\displaystyle \frac{1}{25}(20-18)}\\ & ={\displaystyle \frac{1}{25}(2)=\frac{2}{25}=0,08}\\ \text{c}.& \text{Peluang operator telpon berbicara}\\ & P(2\le X\le 4)\\ & ={\displaystyle {\int }_2^{4}kxdx}\\ & ={\displaystyle {\int }_2^4{\displaystyle \frac{1}{25}xdx}}\\ & ={\displaystyle \frac{1}{25}{\left[{\displaystyle \frac{1}{2}{x}^2}\right]}_2^4}\\ & ={\displaystyle \frac{1}{25}\times \frac{1}{2}(4^2-2^2)}\\ & ={\displaystyle \frac{1}{50}(16-4)}\\ & ={\displaystyle \frac{12}{50}=0,24}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui}x\text{adalah variabel acak kontinu}\\ & \text{yang nilanya berada pada rentang 2 dan 6}\\ & \text{dengan fungsi kepekatannya}f(x)={\displaystyle \frac{1}{20}(x+1).}\\ & \text{Tunjukkan bahwa}P(2<x<6)=\mathbf{\text{1}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}P(2<x<6)& ={\int }_2^6{\displaystyle f(x)dx}\\ & ={\int }_2^6{\displaystyle \frac{1}{20}(x+1)dx}\\ & ={\displaystyle \frac{1}{20}{\int }_2^6{\displaystyle (x+1)dx}}\\ & ={\displaystyle \frac{1}{20}\left({\displaystyle \frac{x^2}{2}+x}\right){|}_2^6}\\ & ={\displaystyle \frac{1}{20}\left({\displaystyle \frac{6^2}{2}+6}\right)-{\displaystyle \frac{1}{20}\left({\displaystyle \frac{2^2}{2}+2}\right)}}\\ & ={\displaystyle \frac{1}{20}(18+6)-{\displaystyle \frac{1}{20}(2+2)}}\\ & ={\displaystyle \frac{1}{20}(24-4)}\\ & ={\displaystyle \frac{20}{20}}\\ & =1(\mathbf{\text{terbukti}})\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}P(2<x<6)& ={\int }_2^6{\displaystyle f(x)dx=1}\\ 1& ={\int }_2^6{\displaystyle \frac{1}{20}(x+1)dx}\\ 1& ={\displaystyle \frac{1}{20}{\int }_2^6{\displaystyle (x+1)dx}}\\ 20& ={\int }_2^6{\displaystyle (x+1)dx}\\ 20& =\left({\displaystyle \frac{x^2}{2}+x}\right){|}_2^6\\ 20& =\left({\displaystyle \frac{6^2}{2}+6}\right)-\left({\displaystyle \frac{2^2}{2}+2}\right)\\ 20& =(18+6)-(2+2)\\ 20& =20(\mathbf{\text{terbukti}})\end{array}\end{array}$.

Sifat-sifat fungsi probabilitas kontinu adalah sebagai berikut

• Modusnya berupa nilai x tertinggi pada interval [a,b]
• Median ($m$) adalah hasil dari persamaan yang melibatan  ${\int }_a^m{\displaystyle f(x)dx={\displaystyle \frac{1}{2}}}$.
• Mean ($\mu$) dirumuskan dengan $\mu ={\int }_a^b{\displaystyle xf(x)dx}$.
• Varian dirumuskan dengan  $var(X)={\int }_a^b{x}^{2}f(x)dx-{\mu }^2$.

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 3.& \text{Diketahui suatu fungsi probabilitas}\\ & f(x)=\{\begin{array}{c}1-{\displaystyle \frac{1}{2}x,\text{pada}[0,2]}\\ \\ 0,\text{pada}x\text{yang lain}\end{array}\\ & \text{a}.\text{Buktian pernyataan di atas benar}\\ & \text{b}.\text{Carilah mean, modus, dan mediannya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Akan ditunjukkan}{\int }_0^2{\displaystyle f(x)dx=1}\\ & {\int }_0^2{\displaystyle f(x)dx={\int }_0^2{\displaystyle (1-{\displaystyle \frac{1}{2}x})dx}}\\ & =(x-{\displaystyle \frac{1}{4}{x}^2}){|}_0^2\\ & =(2-{\displaystyle \frac{1}{4}{.2}^2})-(0-{\displaystyle \frac{1}{4}{.0}^2)}\\ & =(2-1)-(0-0)=1(\mathbf{\text{Terbukti}})\end{array}\\ & \begin{array}{rl}\text{b}.(1)& \text{Mean}=\mu ={\int }_a^{b}x{\displaystyle f(x)dx}\\ & ={\int }_0^{2}x{\displaystyle (1-{\displaystyle \frac{1}{2}x})dx={\int }_0^2{\displaystyle (x-{\displaystyle \frac{1}{2}{x}^2})dx}}\\ & =\left({\displaystyle \frac{1}{2}{x}^2-{\displaystyle \frac{1}{6}{x}^3}}\right){|}_0^2\\ & =(2-1{\displaystyle \frac{2}{6}})-(0)\\ & ={\displaystyle \frac{2}{3}}\end{array}\\ & \begin{array}{rl}\text{b}.(2)& \text{Medus}=\text{nilai maksimum dari}f(x)\\ & f(x)=1-{\displaystyle \frac{1}{2}x,\text{akan maksimum saat}x=0}\\ & \text{maka},f(0)=1-{\displaystyle \frac{1}{2}.0=1-0=1}\end{array}\\ & \begin{array}{rl}\text{b}.(3)& \text{Median}=\text{nilai}m\text{pada}{\int }_0^m{\displaystyle f(x)dx={\displaystyle \frac{1}{2}}}\\ & \text{maka}\\ & {\displaystyle \frac{1}{2}={\int }_0^m{\displaystyle (1-{\displaystyle \frac{1}{2}x})dx=(x-{\displaystyle \frac{1}{4}{x}^2}){|}_0^m}}\\ & \iff {\displaystyle \frac{1}{2}=(m-{\displaystyle \frac{1}{4}{m}^2})-0}\\ & \iff m^2-4m+2=0\\ & \iff m_{1,2}={\displaystyle \frac{4\pm \sqrt{16-4(1)(2)}}{2}}\\ & \iff ={\displaystyle \frac{4\pm 2\sqrt{2}}{2}=2\pm \sqrt{2}}\\ & \iff m_1=2+\sqrt{2}(\text{tidak memenuhi})\\ & \text{lihat batas interval tertutup}[0,2]\\ & \iff m_2=2-\sqrt{2}(\mathbf{\text{memenuhi}})\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diberikan fungsi}\\ & f(x)=\{\begin{array}{c}{\displaystyle \frac{3}{x^2},\text{pada}[1,2]}\\ \\ 0,\text{pada}x\text{yang lain}\end{array}\\ & \text{Selidikilah apakah fungsi tersebut}\\ & \text{fungsi probabilitas atau bukan}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Kita selidiki apakah}0\le f(x)\le 1\\ & f(0)=0,f(1)=3,f(2)={\displaystyle \frac{3}{2^3}=\frac{3}{8}}\\ & \text{Karena terdapat}f(1)=3\ge 1,\text{maka}\\ & \text{telah ditunjuan bahwa fungsi}f(x)\\ & \text{tersebut bukan}\mathbf{\text{fungsi distribusi}}\\ & \mathbf{\text{probabilitas}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kurnia, N., dkk. 2018. Jelajah Matematika SMA Kelas XII Peminatan MIPA. Bogor: YUDHISTIRA.
2. Tasari. Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.