The expecting value E(x) is given by
[tex]E(x)=\frac{\Sigma x_j\cdot t_j}{n}[/tex]where x is the number of problems and t the number of test and n is the total of tests.
By substituting our values, we get
[tex]E(x)=\frac{9\cdot3+34\cdot2+39\cdot6+41\cdot3+50\cdot2+95\cdot4}{3+2+6+3+2+4}[/tex]which gives
[tex]\begin{gathered} E(x)=\frac{932}{20} \\ E(x)=46.6 \end{gathered}[/tex]that is, the expected values is 46.6
