1. Occorre imporre: ${\int }_{-\infty }^{+\infty }f\left(x\right)dx=1\Rightarrow {\int }_{-\infty }^{+\infty }f\left(x\right)dx={\int }_0^{1}adx+{\int }_1^{2}axdx+{\int }_2^{4}a\left(4-x\right)dx=$
   

$=a{\int }_0^{1}dx+a{\int }_1^{2}xdx+4a{\int }_2^{4}dx-a{\int }_2^{4}xdx=a{\left[x\right]}_0^1+a{\left[\frac{x^2}{2}\right]}_1^2+4a{\left[x\right]}_2^4-a{\left[\frac{x^2}{2}\right]}_2^4=$
$=a+2a-\frac{a}{2}+8a-8a+2a=\frac{9}{2}a\Rightarrow a=\frac{2}{9}$

Quindi la distribuzione di probabilità diventa (normalizzata):

  $f\mathrm{(x)}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}& 0\le x<1\\ \frac{2}{9}x& 1\le x<2\\ -\frac{2}{9}x+\frac{8}{9}& 2\le x<4\\ 0& 4\le x\end{array}$

2. Il grafico della f(x) è:

3. La funzione di ripartizione F(x) è definita:   $F\left(x\right)={\int }_{-\infty }^{x}f\left(t\right)dt$

In questo caso:  F(x) $=\{\begin{array}{cc}0& x<0\\ {\int }_0^x\left(\frac{2}{9}\right)dx& 0\le x<1\\ F\left(1\right)+{\int }_1^x\left(\frac{2}{9}x\right)dx& 1\le x<2\\ F\left(2\right)+{\int }_2^x\left(-\frac{2}{9}x+\frac{8}{9}\right)dx& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ F\left(1\right)+{\left[\frac{1}{9}{x}^2\right]}_1^x& 1\le x<2\\ F\left(2\right)+{\left[-\frac{1}{9}{x}^2+\frac{8}{9}x\right]}_2^x& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ \frac{2}{9}+\left[\left(\frac{1}{9}{x}^2\right)-\left(\frac{1}{9}\right)\right]& 1\le x<2\\ F\left(2\right)+\left[\left(-\frac{1}{9}{x}^2+\frac{8}{9}x\right)-\left(-\frac{4}{9}+\frac{16}{9}\right)\right]& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=$

  $=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ \frac{1}{9}{x}^2+\frac{1}{9}& 1\le x<2\\ F\left(2\right)-\frac{1}{9}{x}^2+\frac{8}{9}x-\frac{12}{9}& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ \frac{1}{9}{x}^2+\frac{1}{9}& 1\le x<2\\ \frac{4}{9}+\frac{1}{9}-\frac{1}{9}{x}^2+\frac{8}{9}x-\frac{12}{9}& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ \frac{1}{9}{x}^2+\frac{1}{9}& 1\le x<2\\ -\frac{1}{9}{x}^2+\frac{8}{9}x-\frac{7}{9}& 2\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{9}x& 0\le x<1\\ \frac{1}{9}{x}^2+\frac{1}{9}& 1\le x<2\\ -\frac{1}{9}{x}^2+\frac{8}{9}x-\frac{7}{9}& 2\le x<4\\ 1& 4\le x\end{array}$

Di seguito il grafico della funzione di ripartizione.

4. Calcolo del valore medio:

$m={\int }_{-\infty }^{+\infty }x\cdot f\left(x\right)dx={\int }_0^1\left(\frac{2}{9}\right)xdx+{\int }_1^2\left(\frac{2}{9}x\right)xdx+{\int }_2^4\left(-\frac{2}{9}x+\frac{8}{9}\right)xdx=$

$=\frac{2}{9}{\int }_0^{1}xdx+\frac{2}{9}{\int }_1^2{x}^{2}dx-\frac{2}{9}{\int }_2^4{x}^{2}dx+\frac{8}{9}{\int }_2^{4}xdx=\frac{2}{9}{\left[\frac{x^2}{2}\right]}_0^1+\frac{2}{9}{\left[\frac{x^3}{3}\right]}_1^2-\frac{2}{9}{\left[\frac{x^3}{3}\right]}_2^4+\frac{8}{9}{\left[\frac{x^2}{2}\right]}_2^4=$

$=\frac{2}{9}\left[\frac{1}{2}\right]+\frac{2}{9}\left[\frac{8}{3}-\frac{1}{3}\right]-\frac{2}{9}\left[\frac{64}{3}-\frac{8}{3}\right]+\frac{8}{9}\left[\frac{16}{2}-\frac{4}{2}\right]=\frac{1}{9}+\frac{14}{27}-\frac{112}{27}+\frac{48}{9}=\frac{3+14-112+144}{27}=\frac{49}{27}$

5. Calcolo della varianza:

${\sigma }^2={\int }_{-\infty }^{+\infty }{\left(x-m\right)}^2\cdot f\left(x\right)dx={\int }_0^1\left(\frac{2}{9}\right){\left(x-\frac{49}{27}\right)}^{2}dx+{\int }_1^2\left(\frac{2}{9}x\right){\left(x-\frac{49}{27}\right)}^{2}dx+{\int }_2^4\left(-\frac{2}{9}x+\frac{8}{9}\right){\left(x-\frac{49}{27}\right)}^{2}dx=$

$={\int }_0^1\left(\frac{2}{9}\right)\left(x^2+\frac{2401}{729}-\frac{98}{27}x\right)dx+{\int }_1^2\left(\frac{2}{9}x\right)\left(x^2+\frac{2401}{729}-\frac{98}{27}x\right)dx+{\int }_2^4\left(-\frac{2}{9}x+\frac{8}{9}\right)\left(x^2+\frac{2401}{729}-\frac{98}{27}x\right)dx=$

$={\int }_0^1\left(\frac{2}{9}{x}^2+\frac{4802}{6561}-\frac{196}{243}x\right)dx+{\int }_1^2\left(\frac{2}{9}{x}^3+\frac{4802}{6561}x-\frac{196}{243}{x}^2\right)dx+{\int }_2^4\left(-\frac{2}{9}{x}^3-\frac{4802}{6561}x+\frac{196}{243}{x}^2+\frac{8}{9}{x}^2+\frac{19208}{6561}-\frac{784}{243}x\right)dx=$

$={\int }_0^1\left(\frac{2}{9}{x}^2+\frac{4802}{6561}-\frac{196}{243}x\right)dx+{\int }_1^2\left(\frac{2}{9}{x}^3+\frac{4802}{6561}x-\frac{196}{243}{x}^2\right)dx+{\int }_2^4\left(-\frac{2}{9}{x}^3+\frac{412}{243}{x}^2-\frac{25970}{6561}x+\frac{19208}{6561}\right)dx=$

$={\left[\frac{2}{27}{x}^3+\frac{4802}{6561}x-\frac{98}{243}{x}^2\right]}_0^1+{\left[\frac{1}{18}{x}^4+\frac{2401}{6561}{x}^2-\frac{196}{729}{x}^3\right]}_1^2+{\left[-\frac{1}{18}{x}^4+\frac{412}{729}{x}^3-\frac{25970}{13122}{x}^2+\frac{19208}{6561}x\right]}_2^4=$

$=\frac{2}{27}+\frac{4802}{6561}-\frac{98}{243}+\left[\left(\frac{16}{18}+\frac{9604}{6561}-\frac{1568}{729}\right)-\left(\frac{1}{18}+\frac{2401}{6561}-\frac{196}{729}\right)\right]+\left[\left(-\frac{256}{18}+\frac{26368}{729}-\frac{415520}{13122}+\frac{76832}{6561}\right)-\left(-\frac{16}{18}+\frac{3296}{729}-\frac{103880}{13122}+\frac{38416}{6561}\right)\right]=$
$=\frac{486+4802-2646}{6561}+\left[\frac{11664+19208-28224-729-4802+3528}{13122}\right]+\left[\frac{-186624+474624-415520+153664+11664-59328+103880-76832}{13122}\right]=$
$=\frac{2642}{6561}+\left[\frac{645}{13122}\right]+\left[\frac{5528}{13122}\right]=\frac{5284+645+5528}{13122}=\frac{11457}{13122}=\frac{1273}{1458}$

6. Calcolo della probabilità che la variabile aleatoria x sia al massimo x= 2:

7. Calcolo della probabilità che la variabile aleatoria sia compresa fra 3 e 4:

8. Calcolo della probabilità che la variabile aleatoria sia almeno x= 3: