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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:
\[
F(x) = 
\begin{cases} 
0 & \text{for } x < 1 \\
0.1 & \text{for } x = 1 \\
0.25 & \text{for } x = 2 \\
0.55 & \text{for } x = 3 \\
0.8 & \text{for } x = 4 \\
1 & \text{for } x \geq 5 
\end{cases}
\]

### Step 3: Compute \( P(X > 3) \)
To find \( P(X > 3) \), we use the complement rule:
\[
P(X > 3) = 1 - P(X \leq 3) = 1 - F(3) = 1 - 0.55 = 0.45
\]

Thus, \( P(X > 3) = 0.45 \).

Would you like a detailed explanation of any part? Here are some related questions you might find helpful:

1. How is the cumulative distribution function used in probability?
2. What is the importance of the pmf in discrete random variables?
3. Can the CDF be used to find \( P(X \geq x) \)?
4. How can we derive the pmf from real-world examples?
5. How does the complement rule work in probability theory?

**Tip:** The complement rule is a quick way to compute probabilities for "greater than" or "less than" events when given CDFs or summed probabilities.

Let X be a discrete random variable with pmf given by f(x) = (1/20)(1 + x) for x = 1, 2, 3, 4, 5. Determine the cdf of X, hence compute P(X > 3).

Explore how to compute the cumulative distribution function (CDF) and the probability P(X > 3) for a discrete random variable with a given PMF. Includes step-by-step calculation of probabilities and detailed explanations. Suitable for Grades 11-12.

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