Part a.
The expected value is calculated by multiplying eacn of the possible outcomes by their respective probability and then summing all of those values. In our case, we have
[tex]\begin{gathered} E(X)=\sum_^X_i\times P(X_i) \\ \end{gathered}[/tex]
then
[tex]E(X)=2\times0.05+3\times0.30+4\times0.10+5\times0.15+6\times0.35+7\times0.05[/tex]
which gives
[tex]E(X)=4.6[/tex]
Part b
On the other hand, the sample variance is computed as
[tex]Var(X)=\sum_{j\mathop{=}1}^n(x_i^2*P(X_i)^-\mu^2[/tex]
where μ is the mean value. Given by
[tex]\mu=\frac{2+3+4+5+6+7}{6}=4.5[/tex]
So we get
[tex]\begin{gathered} Var(X)=4*0.05+9*0.3+16*0.1+25*0.15+36*0.35+49*0.05-4.5^2 \\ Var(X)=23.3-20.25 \end{gathered}[/tex]
which gives
[tex]Var(X)=3.05[/tex]
Therefore, the answers are:
a) E(X)= 4.6
b)
Var(X)=3.05
