1. Occorre imporre: ${\int }_{-\infty }^{+\infty }f\left(x\right)dx=1\Rightarrow {\int }_{-\infty }^{+\infty }f\left(x\right)dx={\int }_0^{1}a{x}^{2}dx+{\int }_1^{2}adx-{\int }_2^{3}a\left(x-3\right)dx=$
   

$=a{\int }_0^1{x}^{2}dx+a{\int }_1^{2}dx-a{\int }_2^{3}xdx+3a{\int }_2^{3}dx=a{\left[\frac{x^3}{3}\right]}_0^1+a{\left[x\right]}_1^2-a{\left[\frac{x^2}{2}\right]}_2^3+3a{\left[x\right]}_2^3=$
$=a\frac{1}{3}+a-\frac{9}{2}a+2a+3a=\frac{11}{6}a\Rightarrow a=\frac{6}{11}$

Quindi la distribuzione di probabilità diventa (normalizzata):

  $f\mathrm{(x)}=\{\begin{array}{cc}0& x<0 \\ \frac{6 }{11}{x}^2& 0\le x<1 \\ \frac{6 }{11}& 1\le x<2\\ - \frac{6 }{11}x+ \frac{18}{11}& 2\le x<3\\ 0& 3\le x\end{array}$

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)  $=\{\begin{array}{cc}0& x<0\\ {\int }_0^x\left(\frac{6}{11}{x}^2\right)dx& 0\le x<1\\ F\left(1\right)+{\int }_1^x\left(\frac{6}{11}\right)dx& 1\le x<2\\ F\left(2\right)+{\int }_2^x\left(-\frac{6}{11}x+\frac{18}{11}\right)dx& 2\le x<3\\ F\left(3\right)& 3\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{6}{33}{x}^3& 0\le x<1\\ F\left(1\right)+{\left[\frac{6}{11}x\right]}_1^x& 1\le x<2\\ F\left(2\right)+{\left[-\frac{6}{22}{x}^2+\frac{18}{11}x\right]}_2^x& 2\le x<3\\ F\left(3\right)& 3\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{11}{x}^3& 0\le x<1\\ \frac{6}{33}+\left[\left(\frac{6}{11}x\right)-\left(\frac{6}{11}\right)\right]& 1\le x<2\\ F\left(2\right)+\left[\left(-\frac{6}{22}{x}^2+\frac{18}{11}x\right)-\left(-\frac{12}{11}+\frac{36}{11}\right)\right]& 2\le x<3\\ F\left(3\right)& 3\le x\end{array}=$

  $=\{\begin{array}{cc}0& x<0\\ \frac{2}{11}{x}^3& 0\le x<1\\ \frac{6}{11}x-\frac{4}{11}& 1\le x<2\\ F\left(2\right)-\frac{6}{22}{x}^2+\frac{18}{11}x-\frac{24}{11}& 2\le x<3\\ F\left(3\right)& 3\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{11}{x}^3& 0\le x<1\\ \frac{6}{11}x-\frac{4}{11}& 1\le x<2\\ \frac{12}{11}-\frac{4}{11}-\frac{6}{22}{x}^2+\frac{18}{11}x-\frac{24}{11}& 2\le x<3\\ F\left(3\right)& 3\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{11}{x}^3& 0\le x<1\\ \frac{6}{11}x-\frac{4}{11}& 1\le x<2\\ -\frac{6}{22}{x}^2+\frac{18}{11}x-\frac{16}{11}& 2\le x<3\\ F\left(3\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ \frac{2}{11}{x}^3& 0\le x<1\\ \frac{6}{11}x-\frac{4}{11}& 1\le x<2\\ -\frac{6}{22}{x}^2+\frac{18}{11}x-\frac{16}{11}& 2\le x<3\\ 1& 3\le x\end{array}$

Di seguito il grafico della funzione di ripartizione.

4. Calcolo del valore medio:

$m={\int }_{-\infty }^{+\infty }x\cdot f\left(x\right)dx={\int }_0^1\left(\frac{6}{11}\right)x^{3}dx+{\int }_1^2\left(\frac{6}{11}\right)xdx+{\int }_2^3\left(-\frac{6}{11}x+\frac{18}{11}\right)xdx=$

$=\frac{6}{11}{\int }_0^1{x}^{3}dx+\frac{6}{11}{\int }_1^{2}xdx-\frac{6}{11}{\int }_2^3{x}^{2}dx+\frac{18}{11}{\int }_2^{3}xdx=\frac{6}{11}{\left[\frac{x^4}{4}\right]}_0^1+\frac{6}{11}{\left[\frac{x^2}{2}\right]}_1^2-\frac{6}{11}{\left[\frac{x^3}{3}\right]}_2^3+\frac{18}{11}{\left[\frac{x^2}{2}\right]}_2^3=$

$=\frac{6}{11}\left[\frac{1}{4}\right]+\frac{6}{11}\left[\frac{4}{2}-\frac{1}{2}\right]-\frac{6}{11}\left[\frac{27}{3}-\frac{8}{3}\right]+\frac{18}{11}\left[\frac{9}{2}-\frac{4}{2}\right]=\frac{3}{22}+\frac{9}{11}-\frac{38}{11}+\frac{45}{11}=\frac{3+18-76+90}{22}=\frac{35}{22}$

5. Calcolo della varianza:

${\sigma }^2={\int }_{-\infty }^{+\infty }{\left(x-m\right)}^2\cdot f\left(x\right)dx={\int }_0^1\left(\frac{6}{11}{x}^2\right){\left(x-\frac{35}{22}\right)}^{2}dx+{\int }_1^2\left(\frac{6}{11}\right){\left(x-\frac{35}{22}\right)}^{2}dx+{\int }_2^3\left(-\frac{6}{11}x+\frac{18}{11}\right){\left(x-\frac{35}{22}\right)}^{2}dx=$

$={\int }_0^1\left(\frac{6}{11}{x}^2\right)\left(x^2+\frac{1225}{484}-\frac{35}{11}x\right)dx+{\int }_1^2\left(\frac{6}{11}\right)\left(x^2+\frac{1225}{484}-\frac{35}{11}x\right)dx+{\int }_2^3\left(-\frac{6}{11}x+\frac{18}{11}\right)\left(x^2+\frac{1225}{484}-\frac{35}{11}x\right)dx=$

$={\int }_0^1\left(\frac{6}{11}{x}^4+\frac{7350}{5324}{x}^2-\frac{210}{121}{x}^3\right)dx+{\int }_1^2\left(\frac{6}{11}{x}^2+\frac{7350}{5324}-\frac{210}{121}x\right)dx+{\int }_2^3\left(-\frac{6}{11}{x}^3-\frac{7350}{5324}x+\frac{210}{121}{x}^2+\frac{18}{11}{x}^2+\frac{22050}{5324}-\frac{630}{121}x\right)dx=$

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 2.5 e 3:

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