To complete the table and make it represent a legitimate probability distribution, the sum of all probabilities $P(X = x)$ must equal 1.

Given:

• $P(X = 0) = 0.23$
• $P(X = 5) = 0.23$
• $P(X = 6) = 0.27$

Let the missing values be $P(X = 3) = p_1$ and $P(X = 4) = p_2$.

The total probability should sum to 1:

$0.23 + p_1 + p_2 + 0.23 + 0.27 = 1$

Simplifying, we get:

$p_1 + p_2 + 0.73 = 1$

$p_1 + p_2 = 0.27$

Now, the values of $p_1$ and $p_2$ must be non-negative and add up to 0.27. A simple solution could be:

$p_1 = 0.13 \quad \text{and} \quad p_2 = 0.14$

So, the completed table is:

\hline \text{Value } x \text{ of } X & P(X = x) \\ \hline 0 & 0.23 \\ 3 & 0.13 \\ 4 & 0.14 \\ 5 & 0.23 \\ 6 & 0.27 \\ \hline \end{array}$$ Would you like more details on the calculations or have any other questions? --- Here are some related questions: 1. How do you calculate the mean (expected value) for this discrete probability distribution? 2. What is the cumulative distribution function (CDF) for the variable $$X$$ in this distribution? 3. Can you explain how to calculate the variance of $$X$$ for this distribution? 4. What would change in the distribution if $$P(X = 6)$$ were instead 0.20? 5. How can this distribution be interpreted in a real-world scenario? 6. What other checks can be done to verify if a distribution is valid? 7. If you wanted to add a new possible value $$X = 2$$ to the distribution, how would that affect the probabilities? 8. What are the implications of having two probabilities be exactly equal in a distribution? **Tip:** When distributing probabilities, ensure that the sum of the assigned probabilities equals 1 and that each individual probability is non-negative and does not exceed 1.