To complete the probability distribution for the given discrete random variable $X$, we need to assign probabilities $P(X=x)$ to the values 2 and 5 such that the sum of all probabilities equals 1.

Given:

• $P(X=0) = 0.23$
• $P(X=4) = 0.13$
• $P(X=6) = 0.23$

Let:

• $P(X=2) = p$
• $P(X=5) = q$

The sum of all probabilities for a legitimate probability distribution is:

$P(X=0) + P(X=2) + P(X=4) + P(X=5) + P(X=6) = 1$

Substitute the given values:

$0.23 + p + 0.13 + q + 0.23 = 1$

Simplify:

$0.59 + p + q = 1$

$p + q = 1 - 0.59$

$p + q = 0.41$

So, we need to find values for $p$ and $q$ such that their sum is 0.41. There are multiple pairs that satisfy this condition, but the simplest is to equally distribute the remaining probability:

$p = q = \frac{0.41}{2} = 0.205$

Final Probability Distribution:

• $P(X=0) = 0.23$
• $P(X=2) = 0.205$
• $P(X=4) = 0.13$
• $P(X=5) = 0.205$
• $P(X=6) = 0.23$

Verification:

$0.23 + 0.205 + 0.13 + 0.205 + 0.23 = 1$

This confirms that the distribution is valid.

Would you like a detailed explanation of how to determine probabilities, or have any other questions? Here are some related questions:

1. How do you find the expected value of a discrete random variable?
2. What is the variance and standard deviation of a discrete random variable?
3. Can you explain the concept of cumulative distribution function (CDF)?
4. How do you check if a given set of probabilities is a legitimate distribution?
5. What is the difference between discrete and continuous probability distributions?

Tip: Always ensure that the sum of all probabilities in a distribution is equal to 1.