First, calculate the possible values of X by squaring each face value and subtracting 4:

• For face 1: \(1^2 - 4 = -3\)
• For face 2: \(2^2 - 4 = 0\)
• For face 3: \(3^2 - 4 = 5\)
• For face 6: \(6^2 - 4 = 32\)

The probability distribution table is:

To find \(E(X)\):

\(E(X) = (-3) \cdot \frac{1}{6} + 0 \cdot \frac{1}{2} + 5 \cdot \frac{1}{6} + 32 \cdot \frac{1}{6} = \frac{-3}{6} + \frac{5}{6} + \frac{32}{6} = \frac{34}{6} = \frac{17}{3} \approx 5.67\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{(-3)^2}{6} + \frac{0^2}{2} + \frac{5^2}{6} + \frac{32^2}{6} - \left(\frac{34}{6}\right)^2\)

\(= \frac{9}{6} + \frac{25}{6} + \frac{1024}{6} - \left(\frac{34}{6}\right)^2\)

\(= 144 \left( \frac{1298}{9} \right)\)