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

then

which gives

Part b
On the other hand, the sample variance is computed as

where μ is the mean value. Given by

So we get

which gives

Therefore, the answers are:
a) E(X)= 4.6
b)
Var(X)=3.05