(i) The sum of all probabilities must equal 1:
\(3c + 4c + 5c + 6c = 18c = 1\)
Solving for \(c\):
\(c = \frac{1}{18} = 0.0556\)
(ii) The expected value \(E(X)\) is calculated as:
\(E(X) = \sum x_i P(X = x_i) = 1(3c) + 2(4c) + 3(5c) + 4(6c)\)
\(E(X) = 3c + 8c + 15c + 24c = 50c\)
Substitute \(c = 0.0556\):
\(E(X) = 50 \times 0.0556 = 2.78\)
The variance \(\text{Var}(X)\) is calculated as:
\(\text{Var}(X) = E(X^2) - (E(X))^2\)
\(E(X^2) = 1^2(3c) + 2^2(4c) + 3^2(5c) + 4^2(6c)\)
\(E(X^2) = 3c + 16c + 45c + 96c = 160c\)
\(\text{Var}(X) = 160c - (50c)^2\)
\(\text{Var}(X) = 160c - 2500c^2\)
Substitute \(c = 0.0556\):
\(\text{Var}(X) = 160 \times 0.0556 - 2500 \times (0.0556)^2\)
\(\text{Var}(X) = 1.17\)
(iii) To find \(P(X > E(X))\), we need \(P(X > 2.78)\):
\(P(X > 2.78) = P(X = 3) + P(X = 4) = 5c + 6c = 11c\)
Substitute \(c = 0.0556\):
\(P(X > 2.78) = 11 \times 0.0556 = 0.611\)
