3.2

331

$f\left(x\right)=\begin{cases} \frac{1}{x^4}{,}&kun\ x<-1\ tai\ x>1\\ a{,}&\ kun\ -1\le x\le1 \end{cases}$

Jotta f(x) olisi tiheysfunktio , on oltava f(x)≥0

$x^4\ge0$ kaikilla x. Siten $\frac{1}{x^4}>0$ , kun x < -1 tai x > 1

Lisäksi valitaan a siten, että a ≥ 0, kun -1 ≤ x ≤ 1.

Jotta f(x) olisi tiheysfnktio, on myös oltava

$\int_{-\infty}^{\infty}f\left(x\right)dx=1$

$\int_{-\infty}^{\infty}f\left(x\right)dx=\int_{-\infty}^{-1}\frac{1}{x^4}dx+\int_{-1}^1adx+\int_1^{\infty}\frac{1}{x^4}dx$

$=\lim_{t\rightarrow\infty}\int_t^{-1}x^{-4}dx+\int_{-1}^1adx+\lim_{s\rightarrow\infty}\int_1^sx^{-4}dx$

$=\lim_{t\rightarrow-\infty}\bigg/_{\!\!\!\!\!t}^{-1}-\frac{1}{3x^3}+\bigg/_{\!\!\!\!\!{-1}}^1ax+\lim_{s\rightarrow\infty}\bigg/_{\!\!\!\!\!1}^3-\frac{1}{3x^3}$

$=\lim_{t\rightarrow-\infty}\left(\frac{1}{3}+\frac{1}{3t^2}\right)+\left(a-a\left(a\cdot-1\right)\right)+\lim_{s\rightarrow\infty}\left(-\frac{1}{3s^3}+\frac{1}{3}\right)$
$=\frac{1}{3}+a+a+\frac{1}{3}=\frac{2}{3}+2a$

$\frac{2}{3}+2a=1$
$2a=1-\frac{2}{3}$
$a=\frac{1}{6}$

Tällöin

$f\left(x\right)\begin{cases} \frac{1}{x^4{,}}&kun\ x<-1\ tai\ x>1\\ \frac{1}{6}{,}&kun\ -1\le x\le1 \end{cases}$
$P\left(0\le X\le2\right)=\int_0^2f\left(x\right)dx$
$=\int_0^1\frac{1}{6}dx+\int_1^2\frac{1}{x^4}dx$

$=\frac{11}{24}\approx0{,}46$

332

$f\left(t\right)\begin{cases} 6\left(0{,}25-\left(t-1{,}5\right)^2\right){,}&kun\ 1\le t\le2\\ 0&muualla \end{cases}$

Merkitään todennäköisyyttä, että vähintään 1300 tuntia,

$P\left(X\ge1{,}3\right)$
$P\left(X\ge1{,}3\right)=\int_{1{,}3}^{\infty}f\left(t\right)dt$

$=\int_{1{,}3}^2f\left(t\right)dt+\int_2^{\infty}f\left(t\right)dt$
$=\int_{1{,}3}^26\left(0{,}25-\left(t-1{,}5\right)^2dt+\ \int_2^{\infty}0dt\right)$
$=0{,}784$

Todennäköisyys on noin 0,78

333

$F\left(x\right)=P\left(X\le x\right)=\int_{-\infty}^xf\left(t\right)dt$

$f\left(x\right)\begin{cases} 0{,}&kun\ x\le1\\ \frac{1}{2x\sqrt[]{x}}{,}&kun\ x>1 \end{cases}$

Kun x≤1, niin

$\int_{-\infty}^xf\left(t\right)dt=\int_{-\infty}^x0dt=0$

Kun x>1, niin

$\int_{-\infty}^xf\left(t\right)dt=\ \int_{-\infty}^1f\left(t\right)dt+\int_1^xf\left(t\right)dt=0+\int_1^x\frac{1}{2t\sqrt[]{t}}dt=\bigg/_{\!\!\!\!\!1}^x-\frac{1}{\sqrt[]{t}}=\frac{\sqrt[]{x}-1}{\sqrt[]{x}}=1-\frac{1}{\sqrt[]{x}}$

$F\left(x\right)\begin{cases} 0{,}&kun\ x\le1\\ 1-\frac{1}{\sqrt[]{x}}{,}&kun\ x>1 \end{cases}$

$P\left(X\ge4\right)=1-F\left(4\right)=1-\left(1-\frac{1}{\sqrt[]{4}}\right)=\frac{1}{2}$

$P\left(-1<X\le2{,}25\right)=F\left(2{,}25\right)-F\left(-1\right)=1-\frac{1}{\sqrt[]{2{,}25}}-0=\frac{1}{3}$

335

$F\left(x\right)=P\left(X\le x\right)=\int_{-\infty}^xf\left(t\right)dt$

$f\left(x\right)\begin{cases} \frac{1}{2}-\frac{1}{8}x{,}&kun\ 0\le x\le4\\ 0&mulloin \end{cases}$

Kun x≤0, niin

$\int_{-\infty}^xf\left(t\right)dt=\int_{-\infty}^x0dt$

Kun 0≤x≤4

$\int_{-\infty}^xf\left(t\right)dt=\int_{-\infty}^0f\left(t\right)dt+\int_0^xf\left(t\right)dt$

336

338