The table provided includes the relative frequencies for different outcomes of $X$, and the problem asks us to find $P(X = 9)$ assuming the table represents all possible outcomes.

Table Information:

• $P(X = 3) = 0.04$
• $P(X = 6) = 0.62$
• $P(X = 9) = ?$
• $P(X = 12) = 0.21$

Key Concept:

The sum of all probabilities in a probability distribution must equal 1, because the table represents all possible outcomes.

So: $P(X = 3) + P(X = 6) + P(X = 9) + P(X = 12) = 1$

Calculation:

Substitute the known values: $0.04 + 0.62 + P(X = 9) + 0.21 = 1$

Simplify: $0.87 + P(X = 9) = 1$

Solve for $P(X = 9)$: $P(X = 9) = 1 - 0.87 = 0.13$

Final Answer:

$P(X = 9) = 0.13$

Correct Option:

The correct choice is 0.13.

Would you like further explanation or more examples?

Related Questions:

1. What does it mean for probabilities to form a "distribution"?
2. How do you verify if a table of probabilities is valid?
3. Can probabilities ever be negative? Why or why not?
4. What is the significance of relative frequency in statistics?
5. How do you calculate missing probabilities in incomplete tables?

Tip:

Always check that the sum of probabilities equals 1 when working with probability distributions to ensure the table is valid!