Step 1: Determine the value of $k$. For a cumulative distribution function (CDF) of a discrete random variable, the probability of the last value must be 1. Given $F(5) = \frac{k+2}{4}$. $$F(5) = 1$$ $$\frac{k+2}{4} = 1$$ $$k+2 = 4$$ $$k = 4-2$$ $$k = 2$$ This shows that $k=2$. Step 2: Find $P(1 \le X \le 4)$ in terms of $k$. For a discrete random variable, $P(a \le X \le b)$ is the sum of probabilities for $X$ values from $a$ to $b$. The given table provides the cumulative distribution function $F(x) = P(X \le x)$. Therefore, $P(1 \le X \le 4)$ is equivalent to $P(X \le 4)$, since 1 is the smallest value $X$ can take. $$P(1 \le X \le 4) = F(4)$$ From the table, $F(4) = \frac{k+5}{10}$. $$P(1 \le X \le 4) = \frac{k+5}{10}$$ Using the value of $k=2$: $$P(1 \le X \le 4) = \frac{2+5}{10} = \frac{7}{10}$$ Step 3: Construct the probability mass function (PMF) using $k=2$. The values of $F(x)$ with $k=2$ are: $F(1) = \frac{3}{10}$ $F(2) = \frac{2(2)+3}{20} = \frac{4+3}{20} = \frac{7}{20}$ $F(4) = \frac{2+5}{10} = \frac{7}{10}$ $F(5) = \frac{2+2}{4} = \frac{4}{4} = 1$ Now, calculate the individual probabilities $P(X=x)$: $P(X=1) = F(1) = \frac{3}{10}$ $P(X=2) = F(2) - F(1) = \frac{7}{20} - \frac{3}{10} = \frac{7}{20} - \frac{6}{20} = \frac{1}{20}$ $P(X=4) = F(4) - F(2) = \frac{7}{10} - \frac{7}{20} = \frac{14}{20} - \frac{7}{20} = \frac{7}{20}$ $P(X=5) = F(5) - F(4) = 1 - \frac{7}{10} = \frac{3}{10}$ The PMF is: $x$ | $P(X=x)$ --|----------- 1 | $\frac{3}{10}$ 2 | $\frac{1}{20}$ 4 | $\frac{7}{20}$ 5 | $\frac{3}{10}$ Step 4: Calculate $E(X)$. $$E(X) = \sum x \cdot P(X=x)$$ $$E(X) = 1 \cdot \frac{3}{10} + 2 \cdot \frac{1}{20} + 4 \cdot \frac{7}{20} + 5 \cdot \frac{3}{10}$$ $$E(X) = \frac{3}{10} + \frac{2}{20} + \frac{28}{20} + \frac{15}{10}$$ $$E(X) = \frac{6}{20} + \frac{2}{20} + \frac{28}{20} + \frac{30}{20}$$ $$E(X) = \frac{6+2+28+30}{20} = \frac{66}{20} = \frac{33}{10}$$ $$E(X) = 3.3$$ Step 5: Calculate $Var(X)$. First, calculate $E(X^2)$: $$E(X^2) = \sum x^2 \cdot P(X=x)$$ $$E(X^2) = 1^2 \cdot \frac{3}{10} + 2^2 \