Pembahasan Soal 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 $\int_{-\infty}^{\infty} f(x) \text{ dx} = 1$

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

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

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

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

$\frac{-b}{6} -\frac{-b}{3}= 1\:\:\:\:\:\color{cyan}\text{samakan penyebut}$

$\frac{-b}{6}+\frac{2b}{6} = 1$

$\frac{b}{6} = 1$

$b = 6$

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

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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 $

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

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

$\text{F}(20) = -2 + 4 – 1$

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

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

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

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

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

$x^2 – 40x + 200 = -150$

$x^2 – 40x + 350 = 0$

$x_1 = 27,07\:\:\:\:\:\color{cyan}\text{tidak memenuhi karena di atas 20}$

$x_2 = 12,93$

$Q_{3} = 12,93$

Penyelesaian d

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

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

$x^2 – 40x + 200 = -160$

$x^2 – 40x + 360 = 0$

$x_1 = 26,32\:\:\:\:\:\color{cyan}\text{tidak memenuhi karena di atas 20}$

$x_2 = 13,68$

$\text{P}_{80} = 13,68$