To complete the given table as a probability distribution, the sum of all probabilities $P(x)$ must equal 1. The missing probability value for $P(0)$ can be calculated as follows:

Table Data:

• $P(1) = 0.13$
• $P(2) = 0.09$
• $P(3) = 0.42$
• $P(4) = 0.27$

Total Probability Requirement:

$P(0) + P(1) + P(2) + P(3) + P(4) = 1$

Calculation:

$P(0) = 1 - (P(1) + P(2) + P(3) + P(4))$ $P(0) = 1 - (0.13 + 0.09 + 0.42 + 0.27)$ $P(0) = 1 - 0.91 = 0.09$

Completed Table:

Final Answer:

The missing value is $P(0) = 0.09$.

Would you like further clarification or details about probability distributions?

Related Questions:

1. What are the requirements for a probability distribution?
2. Can probabilities in a distribution be negative?
3. How do you calculate the expected value of $x$ for this table?
4. What is the interpretation of each $P(x)$ value in real-world terms?
5. How would you modify this table if one of the probabilities exceeded 1?

Tip:

Always verify that the sum of all probabilities in a distribution equals exactly 1, as this ensures the table represents a valid probability distribution.