Let’s solve the problem step by step.

Given:

The probability distribution is: $P(X = x) = \frac{kx}{20}, \quad \text{for } x \in \{3, 5, 8, 11\}.$

Part (a): Find the value of $k$

The sum of probabilities for all possible values of $x$ must equal 1: $\sum P(X = x) = 1.$

Substitute $P(X = x) = \frac{kx}{20}$ into the equation: $\frac{k(3)}{20} + \frac{k(5)}{20} + \frac{k(8)}{20} + \frac{k(11)}{20} = 1.$

Simplify: $\frac{k(3 + 5 + 8 + 11)}{20} = 1.$

Evaluate the sum of $x$-values: $3 + 5 + 8 + 11 = 27.$

Substitute this back: $\frac{k(27)}{20} = 1.$

Solve for $k$: $k = \frac{20}{27}.$

Thus, the value of $k$ is: $\boxed{k = \frac{20}{27}}$

Part (b): Find $E(X)$

The expected value $E(X)$ is calculated as: $E(X) = \sum x \cdot P(X = x).$

Substitute $P(X = x) = \frac{kx}{20}$: $E(X) = \sum x \cdot \frac{kx}{20}.$

Substitute $k = \frac{20}{27}$ and compute for all $x \in \{3, 5, 8, 11\}$: $E(X) = \frac{20}{27} \cdot \frac{1}{20} \cdot (3^2 + 5^2 + 8^2 + 11^2).$

First, calculate $3^2 + 5^2 + 8^2 + 11^2$: $3^2 = 9, \quad 5^2 = 25, \quad 8^2 = 64, \quad 11^2 = 121.$

Sum them up: $9 + 25 + 64 + 121 = 219.$

Substitute back: $E(X) = \frac{20}{27} \cdot \frac{1}{20} \cdot 219.$

Simplify: $E(X) = \frac{219}{27}.$

Simplify further: $E(X) = 8.11 \, \text{(approximately)}.$

Thus: $\boxed{E(X) = 8.11}$

Summary:

1. $k = \frac{20}{27}$
2. $E(X) \approx 8.11$

Would you like me to provide further details or clarification?

Follow-Up Questions:

1. How is $E(X)$ interpreted in probability?
2. What would $P(X = x)$ look like for specific values of $x$?
3. How does the choice of $k$ ensure this is a valid probability distribution?
4. Can we calculate $Var(X)$ using $E(X^2)$?
5. What would happen if another value, say $x = 15$, was added to the set?

Tip: Always verify that the sum of probabilities equals 1 to ensure the probability distribution is valid!