a) Calculate the constant C Step 1: For a discrete probability distribution, the sum of all probabilities must equal 1. $$\sum_{x} P(X=x) = 1$$ Step 2: List the probabilities for each value of $x$. For $x=1,2,3$: $P(X=x) = Cx^2$ $P(X=1) = C(1)^2 = C$ $P(X=2) = C(2)^2 = 4C$ $P(X=3) = C(3)^2 = 9C$ For $x=4,5,6$: $P(X=x) = C(7-x)^2$ $P(X=4) = C(7-4)^2 = C(3)^2 = 9C$ $P(X=5) = C(7-5)^2 = C(2)^2 = 4C$ $P(X=6) = C(7-6)^2 = C(1)^2 = C$ Step 3: Sum all probabilities and set equal to 1 to solve for $C$. $$C + 4C + 9C + 9C + 4C + C = 1$$ $$30C = 1$$ $$C = \frac{1}{30}$$ The constant C is $\boxed{\frac{1}{30}}$. b) Calculate the mean of X Step 1: The mean $E(X)$ is calculated using the formula $E(X) = \sum_{x} x \cdot P(X=x)$. Substitute $C = \frac{1}{30}$ into the probabilities: $P(X=1) = \frac{1}{30}$ $P(X=2) = \frac{4}{30}$ $P(X=3) = \frac{9}{30}$ $P(X=4) = \frac{9}{30}$ $P(X=5) = \frac{4}{30}$ $P(X=6) = \frac{1}{30}$ Step 2: Calculate $E(X)$. $$E(X) = (1)\left(\frac{1}{30}\right) + (2)\left(\frac{4}{30}\right) + (3)\left(\frac{9}{30}\right) + (4)\left(\frac{9}{30}\right) + (5)\left(\frac{4}{30}\right) + (6)\left(\frac{1}{30}\right)$$ $$E(X) = \frac{1}{30}(1 + 8 + 27 + 36 + 20 + 6)$$ $$E(X) = \frac{1}{30}(98)$$ $$E(X) = \frac{98}{30} = \frac{49}{15}$$ The mean of X is $\boxed{\frac{49}{15}}$. c) Calculate the variance of X Step 1: The variance $Var(X)$ is calculated using the formula $Var(X) = E(X^2) - (E(X))^2$. First, calculate $E(X^2)$. $$E(X^2) = \sum_{x} x^2 \cdot P(X=x)$$ $$E(X^2) = (1^2)\left(\frac{1}{30}\right) + (2^2)\left(\frac{4}{30}\right) + (3^2)\left(\frac{9}{30}\right) + (4^2)\left(\frac{9}{30}\right) + (5^2)\left(\frac{4}{30}\right) + (6^2)\left(\frac{1}{30}\right)$$ $$E(X^2) = \frac{1}{30}(1 \cdot 1 + 4 \cdot 4 + 9 \cdot 9 + 16 \cdot 9 + 25 \cdot 4 + 36 \cdot 1)$$ $$E(X^2) = \frac{1}{30}(1 + 16 + 81 + 144 + 100 + 36)$$ $$E(X^2) = \frac{1}{30}(378)$$ $$E(X^2) = \frac{378}{30} = \frac{63}{5}$$ Step 2: Calculate $Var(X)$ using $E(X^2)$ and $E(X) = \frac{49}{15}$. $$Var(X) = \frac{63}{5} - \left(\frac{49}{15}\right)^2$$ $$Var(X) = \frac{63}{5} - \frac{2401}{225}$$ $$Var(X) = \frac{63 \times 45}{5 \times 45} - \frac{2401}{225}$$ $$Var(X) = \frac{2835}{225} - \frac{2401}{225}$$ $$Var(X) = \frac{2835 - 2401}{225}$$ $$Var(X) = \frac{434}{225}$$ The variance of X is $\boxed{\frac{434}{225}}$. d) Calculate the mean and variance of Y, where $Y = 3X - 1$ Step 1: Use the properties of expectation for a linear transformation