Let’s break this problem down step by step.

Part 1: Completing the Table

The total relative frequencies should sum to 1. Currently, we have:

$0.3 + 0.1 + 0.1 + 0.1 + 0.4 = 1$

So the table is already complete. There are no missing values.

Part 2: Computing the Relative Frequencies

(a) $P(\{2, 3, 4\})$

The probability of outcomes $\{2, 3, 4\}$ is the sum of their relative frequencies:

$P(\{2, 3, 4\}) = P(2) + P(3) + P(4) = 0.1 + 0.1 + 0.1 = 0.3$

(b) $P(E')$ where $E = \{3, 4\}$

The complement $E'$ includes all outcomes that are not in $E$. So, $E' = \{1, 2, 5\}$.

The probability of $E'$ is the sum of the relative frequencies of these outcomes:

$P(E') = P(1) + P(2) + P(5) = 0.3 + 0.1 + 0.4 = 0.8$

Final Answers:

1. (a): $P(\{2, 3, 4\}) = 0.3$
2. (b): $P(E') = 0.8$

Would you like a further breakdown or have any related questions? Here are some related topics for exploration:

1. What is the probability of $E$ itself ($P(E)$)?
2. What is the sum of $P(E) + P(E')$?
3. How do relative frequencies relate to actual frequencies?
4. Can a relative frequency table have missing probabilities? Why or why not?
5. How can this be visualized with a bar graph?

Tip: Always ensure the probabilities sum to 1 when working with relative frequency distributions!