The number of free throws, and the points he makes has a binomial distribution with the parameters n(amount of free throws) and p(probability to make the free throw).
From the text, those parameters are
[tex]\begin{gathered} n=2 \\ p=54.7\%=0.547 \end{gathered}[/tex]The expectation of the binomial distribution is given by
[tex]\mu=np[/tex]Using our values on this formula, we have
[tex]\mu=2\cdot0.547=1.094[/tex]The expected value of points is 1.094.