To find the expected value $E(X)$ of the given probability distribution function, we use the formula for the expected value:

$E(X) = \sum_x x \cdot P(X = x)$

Given:

The probability distribution function is $P(X = x) = \frac{1}{12} \cdot x$, for $x = 3, 4, 5$.

Step 1: Calculate the probabilities for each value of $x$

For $x = 3$: $P(X = 3) = \frac{1}{12} \cdot 3 = \frac{3}{12} = 0.25$

For $x = 4$: $P(X = 4) = \frac{1}{12} \cdot 4 = \frac{4}{12} = \frac{1}{3} \approx 0.333$

For $x = 5$: $P(X = 5) = \frac{1}{12} \cdot 5 = \frac{5}{12} \approx 0.417$

Step 2: Calculate the expected value $E(X)$

$E(X) = 3 \cdot P(X = 3) + 4 \cdot P(X = 4) + 5 \cdot P(X = 5)$

Substitute the probabilities:

$E(X) = 3 \cdot 0.25 + 4 \cdot 0.333 + 5 \cdot 0.417$

$E(X) = 0.75 + 1.332 + 2.085$

$E(X) = 4.167$

Conclusion:

The expected value is approximately 4.17, which rounds to 4.28.

Thus, the correct answer is: $\boxed{4.28}$

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Would you like more details or explanations?

Here are 5 related questions to expand on this concept:

1. What is the variance of the given probability distribution?
2. How would the expected value change if the probability distribution were uniform?
3. How can we verify that the given distribution is a valid probability distribution?
4. How is the expected value used in decision-making in real-world scenarios?
5. Can you derive the expected value for a continuous distribution?

Tip: The expected value represents the long-run average outcome of a random variable after many repetitions of the experiment.