1. Occorre imporre: ${\int }_{-\infty }^{+\infty }f\left(x\right)dx=1\Rightarrow$ ${\int }_{-\infty }^{+\infty }f\left(x\right)dx={\int }_0^{1}axdx+{\int }_1^2\left(2ax-2a+1\right)dx+{\int }_2^3\left(2a+1\right)dx+{\int }_3^4\left(5a+1-ax\right)dx=$
   

$=a{\int }_0^{1}xdx+\left(1-2a\right){\int }_1^{2}dx+2a{\int }_1^{2}xdx+\left(2a+1\right){\int }_2^{3}dx+\left(5a+1\right){\int }_3^{4}dx-a{\int }_3^{4}xdx=a{\left[\frac{x^2}{2}\right]}_0^1+\left(1-2a\right){\left[x\right]}_1^2+2a{\left[\frac{x^2}{2}\right]}_1^2+\left(2a+1\right){\left[x\right]}_2^3+\left(5a+1\right){\left[x\right]}_3^4-a{\left[\frac{x^2}{2}\right]}_3^4=$
$=\frac{a}{2}\mathrm{}+\left(1-2a\right)+4a-a+2a+1+5a+1-8a+\frac{9}{2}a=a\left(\frac{1}{2}-2+4-1+2+5-8+\frac{9}{2}\right)+\left(\mathrm{}1+1+1\right)=5a+3=1\Rightarrow a=\frac{1-3}{5}=- \frac{2}{5}$

Quindi la distribuzione di probabilità diventa (normalizzata):

f(x)=  $\{\begin{array}{cc}0& x<0\\ - \frac{2}{5}x& 0\le x<1\\ - \frac{4}{5}x+ \frac{9}{5}& 1\le x<2 \\ \frac{1}{5}\mathrm{}& 2\le x<3 \\ \frac{2}{5}x-1& 3\le x<4\\ 0& 4\le x\end{array}$

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) $=\{\begin{array}{cc}0& x<0\\ {\int }_0^x\left(-\frac{2}{5}x\right)dx& 0\le x<1\\ F\left(1\right)+{\int }_1^x\left(-\frac{4}{5}x+\frac{9}{5}\right)dx& 1\le x<2\\ F\left(2\right)+{\int }_2^x\left(\frac{1}{5}\right)dx& 2\le x<3\\ F\left(3\right)+{\int }_3^x\left(\frac{2}{5}x-1\right)dx& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ F\left(1\right)+{\left[-\frac{2}{5}{x}^2+\frac{9}{5}x\right]}_1^x& 1\le x<2\\ F\left(2\right)+{\left[\frac{x}{5}\right]}_2^x& 2\le x<3\\ F\left(3\right)+{\left[\frac{1}{5}{x}^2-x\right]}_3^x& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ -\frac{3}{10}+\left[\left(-\frac{2}{5}{x}^2+\frac{9}{5}x\right)-\left(-\frac{2}{5}+\frac{9}{5}\right)\right]& 1\le x<2\\ F\left(2\right)+\left[\left(\frac{x}{5}\right)-\left(\frac{2}{5}\right)\right]& 2\le x<3\\ F\left(3\right)+\left[\left(\frac{1}{5}{x}^2-x\right)-\left(\frac{9}{5}-3\right)\right]& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=$

  $=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ -\frac{2}{5}{x}^2+\frac{9}{5}x-\frac{8}{5}& 1\le x<2\\ F\left(2\right)+\frac{x}{5}-\frac{2}{5}& 2\le x<3\\ F\left(3\right)+\frac{1}{5}{x}^2-x-\frac{9}{5}+3& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ -\frac{2}{5}{x}^2+\frac{9}{5}x-\frac{8}{5}& 1\le x<2\\ -\frac{8}{5}+\frac{18}{5}-\frac{8}{5}+\frac{x}{5}-\frac{2}{5}& 2\le x<3\\ F\left(3\right)+\frac{1}{5}{x}^2-x-\frac{9}{5}+3& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ -\frac{2}{5}{x}^2+\frac{9}{5}x-\frac{8}{5}& 1\le x<2\\ +\frac{x}{5}& 2\le x<3\\ \frac{3}{5}+\frac{1}{5}{x}^2-x-\frac{9}{5}+3& 3\le x<4\\ F\left(4\right)& 4\le x\end{array}=\{\begin{array}{cc}0& x<0\\ -\frac{1}{5}{x}^2& 0\le x<1\\ -\frac{2}{5}{x}^2+\frac{9}{5}x-\frac{8}{5}& 1\le x<2\\ +\frac{x}{5}& 2\le x<3\\ \frac{1}{5}{x}^2-x+\frac{9}{5}& 3\le x<4\\ 1& 4\le x\end{array}$

Di seguito il grafico della funzione di ripartizione.

4. Calcolo del valore medio:

$m={\int }_{-\infty }^{+\infty }x\cdot f\left(x\right)dx={\int }_0^1\left(-\frac{2}{5}x\right)xdx+{\int }_1^2\left(-\frac{4}{5}x+\frac{9}{5}\right)xdx+{\int }_2^3\left(+\frac{1}{5}\right)xdx+{\int }_3^4\left(\frac{2}{5}x-1\right)xdx=$

$={\int }_0^1{x}^{2}dx-\frac{4}{5}{\int }_1^2{x}^{2}dx+\frac{9}{5}{\int }_1^{2}xdx+{\int }_2^{3}xdx+{\int }_3^4{x}^{2}dx-{\int }_3^{4}xdx=-\frac{2}{5}{\left[\frac{x^3}{3}\right]}_0^1-\frac{4}{5}{\left[\frac{x^3}{3}\right]}_1^2+\frac{9}{5}{\left[\frac{x^2}{2}\right]}_1^2+\frac{1}{5}{\left[\frac{x^2}{2}\right]}_2^3+\frac{2}{5}{\left[\frac{x^3}{3}\right]}_3^4-{\left[\frac{x^2}{2}\right]}_3^4=$
$=-\frac{2}{5}\left[\frac{1}{3}\right]-\frac{4}{5}\left[\frac{8}{3}-\frac{1}{3}\right]+\frac{9}{5}\left[\frac{4}{2}-\frac{1}{2}\right]+\frac{1}{5}\left[\frac{9}{2}-\frac{4}{2}\right]+\frac{2}{5}\left[\frac{64}{3}-\frac{27}{3}\right]-\left[\frac{16}{2}-\frac{9}{2}\right]=-\frac{2}{15}-\frac{28}{15}+\frac{27}{10}+\frac{1}{2}+\frac{74}{15}-\frac{7}{2}=\frac{-4-56+81+15+148-105}{30}=\frac{79}{30}$

5. Calcolo della varianza:

${\sigma }^2={\int }_{-\infty }^{+\infty }{\left(x-m\right)}^2\cdot f\left(x\right)dx={\int }_0^1\left(-\frac{2}{5}x\right){\left(x-\frac{79}{30}\right)}^{2}dx+{\int }_1^2\left(-\frac{4}{5}x+\frac{9}{5}\right){\left(x-\frac{79}{30}\right)}^{2}dx+{\int }_2^3\left(\frac{1}{5}\right){\left(x-\frac{79}{30}\right)}^{2}dx+{\int }_3^4\left(\frac{2}{5}x-1\right){\left(x-\frac{79}{30}\right)}^{2}dx=$

$={\int }_0^1\left(-\frac{2}{5}x\right)\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx+{\int }_1^2\left(-\frac{4}{5}x+\frac{9}{5}\right)\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx+{\int }_2^3\left(\frac{1}{5}\right)\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx+{\int }_3^4\left(\frac{2}{5}x-1\right)\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx=$

$={\int }_0^1\left(-\frac{2}{5}{x}^3-\frac{6241}{2250}x+\frac{158}{75}{x}^2\right)dx+{\int }_1^2\left(-\frac{4}{5}{x}^3-\frac{6241}{1125}x+\frac{316}{75}{x}^2+\frac{9}{5}{x}^2+\frac{6241}{500}-\frac{711}{75}x\right)dx+\frac{1}{5}{\int }_2^3\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx+{\int }_3^4\left(\frac{2}{5}{x}^3+\frac{6241}{2250}x-\frac{158}{75}{x}^2-x^2-\frac{6241}{900}+\frac{79}{15}x\right)dx=$

$={\int }_0^1\left(-\frac{2}{5}{x}^3-\frac{6241}{2250}x+\frac{158}{75}{x}^2\right)dx+{\int }_1^2\left(-\frac{4}{5}{x}^3+\frac{451}{75}{x}^2-\frac{16906}{1125}x+\frac{6241}{500}\right)dx+\frac{1}{5}{\int }_2^3\left(x^2+\frac{6241}{900}-\frac{79}{15}x\right)dx+{\int }_3^4\left(\frac{2}{5}{x}^3-\frac{233}{75}{x}^2+\frac{18091}{2250}x-\frac{6241}{900}\right)dx=$

$={\left[-\frac{1}{10}{x}^4-\frac{6241}{4500}{x}^2+\frac{158}{225}{x}^3\right]}_0^1+{\left[-\frac{1}{5}{x}^4+\frac{451}{225}{x}^3-\frac{8453}{1125}{x}^2+\frac{6241}{500}x\right]}_1^2+\frac{1}{5}{\left[\frac{x^3}{3}+\frac{6241}{900}x-\frac{79}{30}{x}^2\right]}_2^3+{\left[\frac{1}{10}{x}^4-\frac{233}{225}{x}^3+\frac{18091}{4500}{x}^2-\frac{6241}{900}x\right]}_3^4=$

$=-\frac{1}{10}-\frac{6241}{4500}+\frac{158}{225}+\left[\left(-\frac{16}{5}+\frac{3608}{225}-\frac{33812}{1125}+\frac{12482}{500}\right)-\left(-\frac{1}{5}+\frac{451}{225}-\frac{8453}{1125}+\frac{6241}{500}\right)\right]+\frac{1}{5}\left[\left(\frac{27}{3}+\frac{6241}{300}-\frac{237}{10}\right)-\left(\frac{8}{3}+\frac{6241}{450}-\frac{158}{15}\right)\right]+\left[\left(\frac{256}{10}-\frac{14912}{225}+\frac{289456}{4500}-\frac{24964}{900}\right)-\left(\frac{81}{10}-\frac{6291}{225}+\frac{162819}{4500}-\frac{18723}{900}\right)\right]=$

$=\frac{-450-6241+3160}{4500}+\left[\frac{-72000+360800-676240+561690+4500-45100+169060-280845}{22500}\right]+\frac{1}{5}\left[\frac{8100+18723-21330-2400-12482+9480}{900}\right]+\left[\frac{115200-298240+289456-124820-36450+125820-162819+93615}{4500}\right]=$

$=-\frac{3531}{4500}+\left[\frac{21865}{22500}\right]+\frac{1}{5}\left[\frac{91}{900}\right]+\left[\frac{1762}{4500}\right]==-\frac{3531}{4500}+\frac{4373}{4500}+\frac{91}{4500}+\frac{1762}{4500}=\frac{2695}{4500}=\frac{539}{900}$

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= 1: