The problem presents a discrete random variable $X$ with a given probability mass function (pmf) $f(x)$. It asks for the cumulative distribution function (CDF) and to compute $P(X > 3)$.

Step 1: PMF $f(x)$

The pmf is provided as: $f(x) = \frac{1}{20}(1+x) \quad \text{for} \quad x = 1, 2, 3, 4, 5$ $f(x) = 0 \quad \text{elsewhere}.$

We can calculate the individual probabilities:

• For $x = 1$, $f(1) = \frac{1}{20}(1+1) = \frac{2}{20} = 0.1$.
• For $x = 2$, $f(2) = \frac{1}{20}(1+2) = \frac{3}{20} = 0.15$.
• For $x = 3$, $f(3) = \frac{1}{20}(1+3) = \frac{4}{20} = 0.2$.
• For $x = 4$, $f(4) = \frac{1}{20}(1+4) = \frac{5}{20} = 0.25$.
• For $x = 5$, $f(5) = \frac{1}{20}(1+5) = \frac{6}{20} = 0.3$.

Step 2: CDF of $X$

The cumulative distribution function $F(x) = P(X \leq x)$ is computed by summing the probabilities up to $x$.

• $F(1) = f(1) = 0.1$
• $F(2) = f(1) + f(2) = 0.1 + 0.15 = 0.25$
• $F(3) = f(1) + f(2) + f(3) = 0.1 + 0.15 + 0.2 = 0.55$
• $F(4) = f(1) + f(2) + f(3) + f(4) = 0.1 + 0.15 + 0.2 + 0.25 = 0.8$
• $F(5) = f(1) + f(2) + f(3) + f(4) + f(5) = 0.1 + 0.15 + 0.2 + 0.25 + 0.3 = 1.0$

Thus, the CDF is: