To find the expected value of this probability distribution, we can use this expression:
[tex]\begin{gathered} E(x)=\sum ^{}_{}x\cdot p \\ E(x)=(2)(\frac{1}{36})+(3)(\frac{1}{18})+(4)(\frac{1}{12})+(5)(\frac{1}{9})+(6)(\frac{5}{36})+(7)(\frac{1}{6})+(8)(\frac{5}{36})+(9)(\frac{1}{9})+(10)(\frac{1}{12})+(11)(\frac{1}{18})+(12)(\frac{1}{36}) \end{gathered}[/tex]
Solving this E(x) is equal to:
[tex]E(x)=7[/tex]
