To create a legitimate probability distribution for the discrete random variable $X$, the probabilities $P(X = x)$ must satisfy two conditions:

1. Each probability $P(X = x)$ must be between 0 and 1.
2. The sum of all probabilities must equal 1.

You provided the following values:

• $P(X = -4) = 0.30$
• $P(X = -1)$ (to be determined)
• $P(X = 4) = 0.23$
• $P(X = 5) = 0.13$
• $P(X = 6)$ (to be determined)

Let's denote the unknown probabilities as follows:

• $P(X = -1) = p_1$
• $P(X = 6) = p_2$

The sum of all probabilities must be:

$P(X = -4) + p_1 + P(X = 4) + P(X = 5) + p_2 = 1$

Substituting the known values:

$0.30 + p_1 + 0.23 + 0.13 + p_2 = 1$

Simplifying this equation:

$0.66 + p_1 + p_2 = 1$

$p_1 + p_2 = 0.34$

The values of $p_1$ and $p_2$ should be non-negative and must add up to 0.34. One possible solution is to equally distribute the remaining probability:

$p_1 = p_2 = \frac{0.34}{2} = 0.17$

Thus, the complete probability distribution is:

• $P(X = -4) = 0.30$
• $P(X = -1) = 0.17$
• $P(X = 4) = 0.23$
• $P(X = 5) = 0.13$
• $P(X = 6) = 0.17$

This satisfies the requirement for a legitimate probability distribution.

Would you like more details on this solution or have any questions?

Here are 8 follow-up questions:

1. Can you think of any other values for $p_1$ and $p_2$ that would also satisfy the conditions?
2. Why must the sum of all probabilities equal 1?
3. How would you interpret the probability $P(X = x) = 0.23$?
4. What are some examples of real-world situations where such a probability distribution might be used?
5. How would the distribution change if $P(X = -4)$ were increased?
6. What would happen if one of the probabilities were negative?
7. Can this probability distribution be represented graphically?
8. How would the calculation change if another value of $X$ were added?

Tip: When creating a probability distribution, always double-check that the sum of all probabilities equals 1.