Questions: Your answer is incorrect. Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, 2, 4, 5, and 6. Value x of X P(X=x) -3 0.22 2 0.22 4 5 0.26

Solution Steps

A legitimate probability distribution must satisfy two conditions:

1. Each probability \( P(X = x) \) must be between 0 and 1, inclusive: \( 0 \leq P(X = x) \leq 1 \).
2. The sum of all probabilities must equal 1: \( \sum P(X = x) = 1 \).

The given probabilities are:

• \( P(X = -3) = 0.22 \)
• \( P(X = 2) = 0.22 \)
• \( P(X = 5) = 0.26 \)

Sum of the given probabilities: \[ 0.22 + 0.22 + 0.26 = 0.70 \]

Since the total probability must equal 1, the missing probability \( P(X = 4) \) is: \[ P(X = 4) = 1 - 0.70 = 0.30 \]

The completed probability distribution is: \[ \begin{tabular}{|c|c|} \hline Value \( x \) of \( X \) & \( P(X = x) \) \\ \hline -3 & 0.22 \\ \hline 2 & 0.22 \\ \hline 4 & 0.30 \\ \hline 5 & 0.26 \\ \hline \end{tabular} \] All probabilities are between 0 and 1, and their sum is 1, so this is a legitimate probability distribution.

Final Answer