Pembahasan Variabel Acak Kontinu

\(f(x)\) adalah suatu fungsi padat peluang jika memenuhi:

1. \(f(x) \geqslant 0, \text{ untuk setiap nilai x}\)
2. \(\int_{-\infty}^{\infty} f(x) \text{ dx} = 1\)

\(f(x)\) selalu bernilai positif untuk setiap nilai x, misalkan kita ambil:

\(x = -1 \longrightarrow f(-1) = 0\)

\(x = 0 \longrightarrow f(0) = -\frac{3}{4}(0)(0 -2) = 0\)

\(x = 1 \longrightarrow f(2) = -\frac{3}{4}(1)(1 -2) = \frac{3}{4}\)

\(\text{dan seterusnya…}\)

Selanjutnya, akan ditunjukkan \(\int_{0}^{2}-\frac{3}{4}x(x-2)\text{ dx} = 1\)

\( \int_{0}^{2} -\frac{3}{4}x^2 + \frac{3}{2}x \text{ dx}\)

\(\left. (-\frac{1}{4}x^3 + \frac{3}{4}x^2) \right |_{0}^{2}\)

\(-\frac{1}{4}(2)^3 + \frac{3}{4}(2)^2 – 0\)

\(-2 + 3\)

\(1\)

Karena \(f(x)\geqslant 0, \text{ untuk setiap x}\) dan  \(\int_{0}^{2}(-\frac{3}{4}x(x-2))\text{ dx} = 1\), maka \(f(x)\) adalah suatu fungsi padat peluang

Penyelesaian (a)

Karena \(f(x)\) merupakan fungsi padat peluang maka \(\int_{-\infty}^{\infty} f(x) \text{ dx} = 1\)

\(\int_{0}^{1} a(6-2x) \text{ dx} = 1\)

\(\int_{0}^{1} 6a – 2ax \text{ dx} = 1\)

\(\left. 6ax – ax^2 \right |_{0}^{1}= 1\)

\(6a(1) – a(1)^2 – 0  = 1\)

\(6a – a = 1\)

\(5a = 1\)

\(a = \frac{1}{5}\)

Penyelesaian (b)

\(\text{P}(X \leqslant 0,8) = \int_{0}^{0,8} \frac{1}{5}(6-2x) \text{ dx}\)

\(\int_{0}^{0,8} \frac{6}{5} – \frac{2}{5}x \text{ dx}\)

\(\left.(\frac{6}{5}x – \frac{1}{5}x^2) \right |_{0}^{0,8} \)

\(\frac{6}{5}(0,8) – \frac{1}{5}(0,8)^2  – 0\)

\(0,96 – 0,128\)

\(0,832\)

Penyelesaian (a)

Karena \(f(x)\) merupakan fungsi padat peluang maka \(\int_{-\infty}^{\infty} f(x) \text{ dx} = 1\)

\(\int_{3}^{6} \frac{b}{x^2}\text{ dx} = 1\)

\(\int_{3}^{6} b\cdot x^{-2}\text{ dx} = 1\)

\(\left.\frac{b}{-1}\cdot x^{-1}\right |_{3}^{6}= 1\)

\(\left.\frac{-b}{x}\right |_{3}^{6}= 1\)

\(\frac{-b}{6} -\frac{-b}{3}= 1\:\:\:\:\:\color{blue}\text{samakan penyebut}\)

\(\frac{-b}{6}+\frac{2b}{6} = 1\)

\(\frac{b}{6} = 1\)

\(b = 6\)

 

Penyelesaian (b)

\(\text{P}(X \geqslant 4) = \int_{4}^{6} \frac{6}{x^2} \text{ dx}\)

\(\int_{4}^{6} 6\cdot x^{-2}\text{ dx}\)

\(\left.\frac{6}{-1}x^{-1}\right |_{4}^{6}\)

\(\left.\frac{-6}{x}\right |_{4}^{6}\)

\(\frac{-6}{6} – \frac{-6}{4}\)

\(- 1 + \frac{3}{2}\)

\(\frac{1}{2}\)

 

Penyelesaian (c)

\(\text{P}(3\leqslant X \leqslant 5) = \int_{3}^{5} \frac{6}{x^2} \text{ dx}\)

\(\int_{3}^{5} 6\cdot x^{-2}\text{ dx}\)

\(\left.\frac{6}{-1}x^{-1}\right |_{3}^{5}\)

\(\left.\frac{-6}{x}\right |_{3}^{5}\)

\(\frac{-6}{5} – (\frac{-6}{3})\)

\(\frac{-6}{5} + 2\)

\(0,8\)

a. Rata-rata

\(\text{Rata-rata}  = \int_{-\infty}^{\infty} x\cdot \text{f(x)} \text{ dx}\)

\(\int_{-\infty}^{0}\text{ dx}  + \int_{0}^{\infty} 0,2x\cdot e^{-0,2x} \text{ dx}\)

\(\left. x \right |_{-\infty}^{0}+ \int_{0}^{\infty} 0,2x\cdot e^{-0,2x} \text{ dx}\)

\(0 + \int_{0}^{\infty} 0,2x\cdot e^{-0,2x} \text{ dx}\)

\(\text{Gunakan integral parsial}\)

\(\int u\cdot \text{ dv} = uv – \int v \cdot \text{ du}\)

\(\text{misalkan: }\)

\(\text{u } = 0,2x  \longrightarrow \text{du} = 0,2 \text{ dx}\)

\(\text{dv }  = e^{-0,2x} \text{ dx}\longrightarrow \text{v } = \int e^{-0,2x} \text{ dx}\)

\(\text{v }  = -\frac{1}{0,2}e^{-0,2x}\)

 

\(\int_{0}^{\infty} 0,2x\cdot e^{-0,2x} \text{ dx} \)

\(\left. 0,2x\cdot (-\frac{1}{0,2}e^{-0,2x})\right |_{0}^{\infty}-\int_{0}^{\infty}  (-\frac{1}{0,2}e^{-0,2x})\cdot 0,2 \text{ dx}\)

\(\left.(-xe^{-0,2x})\right |_{0}^{\infty} +\int_{0}^{\infty} e^{-0,2x}\text{ dx}\)

\(0 + \left.\frac{1}{-0,2}e^{-0,2x}\right|_{0}^{\infty} \)

\(0 – (-\frac{10}{2})\)

\(5\)

\(\text{Jadi nilai rata-ratanya adalah 5 }\)

 

b. Median

\(\text{Median } = P(X \leq m)= 0,5\)

\(\int_{0}^{m} 0,2 e^{-0,2x} \text{ dx} = 0,5\)

\(\left. – e^{-0,2x}\right |_{0}^{m}  = 0,5\)

\(- e^{-0,2m} – (-1) = 0,5\)

\(- e^{-0,2m}  = 0,5 – 1\)

\(- e^{-0,2m} = -0,5\)

\(e^{-0,2m} = 0,5\)

\(^e \log 0,5 = -0,2m\)

\(-0,693 = -0,2m\)

\(m = \frac{0,693}{0,2}\)

\(m = 3,465\)

\(\text{Jadi nilai mediannya 3,465}\)

 

c. Modus

Fungsi padat peluang \(f(x) = 0,2 e^{-0,2x}\) pada interval \(x \geqslant 0\) memiliki nilai maksimum 0,2 untuk \(x = 0\), sehingga nilai modusnya adalah 0.

 

c. Ragam/variansi

\(\text{Ragam} = \int_{a}^{b} x^2\cdot \text{f(x)} \text{ dx} – (\int_{a}^{b} x\cdot \text{f(x)} \text{ dx} )^2\)

\(\int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx} – (\int_{0}^{\infty} 0,2x e^{-0,2x} \text{ dx} )^2\)

\(\int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx} – (5)^2\)

\(\int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx} – 25\)

 

\(\text{Menghitung } \int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx}\)

\(\text{Gunakan integral parsial}\)

\(\int u\cdot \text{ dv} = uv – \int v \cdot \text{ du}\)

\(\text{misalkan: }\)

\(\text{u } = 0,2x^2  \longrightarrow \text{du} = 0,4x \text{ dx}\)

\(\text{dv }  = e^{-0,2x} \text{ dx}\longrightarrow \text{v } = \int e^{-0,2x} \text{ dx}\)

\(\text{v }  = -\frac{1}{0,2}e^{-0,2x}\)

 

\(\int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx}\)

\(\left. 0,2x^2\cdot (-\frac{1}{0,2}e^{-0,2x})\right |_{0}^{\infty}-\int_{0}^{\infty}  (-\frac{1}{0,2}e^{-0,2x})\cdot 0,4x \text{ dx}\)

\(\left. (-x^2\cdot e^{-0,2x})\right |_{0}^{\infty}+ \int_{0}^{\infty} 2x\cdot e^{-0,2x} \text{ dx}\)

\(0 + \int_{0}^{\infty} 2x\cdot e^{-0,2x} \text{ dx}\)

\(\int_{0}^{\infty} 2x\cdot e^{-0,2x} \text{ dx}\)

\(\text{Menghitung }\int_{0}^{\infty} 2x\cdot e^{-0,2x} \text{ dx}\)

\(\int u\cdot \text{ dv} = uv – \int v \cdot \text{ du}\)

\(\text{misalkan: }\)

\(\text{u } = 2x \longrightarrow \text{du} = 2\text{ dx}\)

\(\text{dv }  = e^{-0,2x} \text{ dx}\longrightarrow \text{v } = \int e^{-0,2x} \text{ dx}\)

\(\text{v }  = -\frac{1}{0,2}e^{-0,2x}\)

 

\(\int_{0}^{\infty} 2x\cdot e^{-0,2x} \text{ dx}\)

\(\left. 2x\cdot (-\frac{1}{0,2}e^{-0,2x})\right |_{0}^{\infty}-\int_{0}^{\infty}  (-\frac{1}{0,2}e^{-0,2x})\cdot 2 \text{ dx}\)

\(0 + 10\int_{0}^{\infty} e^{-0,2x}\text{ dx}\)

\(\left. (10 \cdot \frac{1}{-0,2} e^{-0,2x}) \right |_{0}^{\infty}\)

\(0 + 50\)

\(50\)

\(\text{Jadi, }\int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx} = 50\)

\(\text{Ragam} = \int_{0}^{\infty} 0,2x^2 e^{-0,2x} \text{ dx} – 25\)

\(\text{Ragam} = 50 – 25\)

\(\text{Ragam} = 25\)

 

d. Simpangan Baku

\(\text{Simpangan baku} = \sqrt{\text{variansi}}\)

\(\text{Simpangan baku} = \sqrt{25}\)

\(\text{Simpangan baku} = 5\)

Penyelesaian (a)

Suatu fungsi distribusi kumulatif, \(\text{F}(x)\) bernilai sama dengan \(\text{P}(X \leqslant x)\)

Untuk interval \(0\leqslant  x \leqslant 10\)

\(\text{F}(x) = \int_{0}^{x} \frac{1}{100}x \text{ dx}\)

\(\left.  \frac{1}{200}x^2 \right |_{0}^{x}\)

\(\frac{1}{200}x^2 – 0\)

\(\frac{1}{200}x^2 \)

\(\text{F}(10) = \frac{1}{200}(10)^2 \)

\(\text{F}(10) = \frac{1}{2}\)

Untuk interval \(10 <  x \leqslant 20\)

\(\text{F}(x) = \text{F}(10) + \int_{10}^{x} -\frac{1}{100}(x – 20) \text{ dx} \)

\(\frac{1}{2} + \int_{10}^{x} -\frac{1}{100}x + \frac{1}{5} \text{ dx}\)

\(\left. \frac{1}{2} + (-\frac{1}{200}x^2 + \frac{1}{5}x)\right |_{10}^{x}\)

\(\frac{1}{2} -\frac{1}{200}x^2 + \frac{1}{5}x – (-\frac{1}{200}(10)^2 + \frac{1}{5}(10)) \)

\(\frac{1}{2} -\frac{1}{200}x^2 + \frac{1}{5}x  + \frac{1}{2} – 2 \)

\(-\frac{1}{200}x^2 + \frac{1}{5}x  – 1\)

\(\text{F}(20) = -\frac{1}{200}(20)^2 + \frac{1}{5}(20)  – 1\)

\(\text{F}(20) = -2 + 4 – 1\)

\(\text{F}(20) = 1\)

 

$$\bbox[yellow, 5px]{\text{F}(x) = \begin{cases} 0 &, x < 0\\ \frac{1}{200}x^2 &, 0\leqslant  x \leqslant 10\\ -\frac{1}{200}x^2 + \frac{1}{5}x  – 1 &,  10 <  x \leqslant 20 \\1 &, x > 20\end{cases}}$$

 

Penyelesaian b

\(Q_{1} = \text{F}(x) = 0.25\)

\(\frac{1}{200}x^2 = 0.25\)

\(x^2 = 50\)

\(x = \pm \sqrt{50}\)

\(x = + 5\sqrt{2}\)

\(Q_{1} =  5\sqrt{2}\)

 

Penyelesaian c

\(Q_{3} = \text{F}(x) = 0,75\)

\(-\frac{1}{200}x^2 + \frac{1}{5}x  – 1 = 0,75\:\:\:\:\:\color{blue}\text{kalikan kedua ruas dengan -200}\)

\(x^2 – 40x + 200 = -150\)

\(x^2 – 40x + 350 = 0\)

\(x_1 = 27,07\:\:\:\:\:\color{blue}\text{tidak memenuhi karena di atas 20}\)

\(x_2 = 12,93\)

\(Q_{3} =  12,93\)

 

Penyelesaian d

\(\text{P}_{80} = \text{F}(x) = 0,80\)

\(-\frac{1}{200}x^2 + \frac{1}{5}x  – 1 = 0,80\:\:\:\:\:\color{blue}\text{kalikan kedua ruas dengan -200}\)

\(x^2 – 40x + 200 = -160\)

\(x^2 – 40x + 360 = 0\)

\(x_1 = 26,32\:\:\:\:\:\color{blue}\text{tidak memenuhi karena di atas 20}\)

\(x_2 = 13,68\)

\(\text{P}_{80} =  13,68\)