Math Problem Statement
Solution
The problem is asking to fill in the missing probabilities in a table to create a legitimate probability distribution for the discrete random variable .
We know that for a valid probability distribution:
- All probabilities must be non-negative.
- The sum of the probabilities must equal 1.
Given table:
\hline \text{Value } x \text{ of } X & P(X = x) \\ \hline -3 & 0.25 \\ 0 & ? \\ 4 & ? \\ 5 & 0.24 \\ 6 & 0.18 \\ \hline \end{array}$$ ### Step 1: Add the known probabilities. $$P(X = -3) = 0.25, \quad P(X = 5) = 0.24, \quad P(X = 6) = 0.18$$ Summing these known probabilities: $$0.25 + 0.24 + 0.18 = 0.67$$ ### Step 2: Use the fact that the total probability must equal 1. $$P(X = -3) + P(X = 0) + P(X = 4) + P(X = 5) + P(X = 6) = 1$$ Substituting the known probabilities: $$0.67 + P(X = 0) + P(X = 4) = 1$$ $$P(X = 0) + P(X = 4) = 1 - 0.67 = 0.33$$ ### Step 3: Assign values to $$P(X = 0)$$ and $$P(X = 4)$$. Since $$P(X = 0) + P(X = 4) = 0.33$$, there are various ways to distribute this. If no further constraints are given, one possible distribution is: $$P(X = 0) = 0.15, \quad P(X = 4) = 0.18$$ Thus, the completed table could be: $$\begin{array}{|c|c|} \hline \text{Value } x \text{ of } X & P(X = x) \\ \hline -3 & 0.25 \\ 0 & 0.15 \\ 4 & 0.18 \\ 5 & 0.24 \\ 6 & 0.18 \\ \hline \end{array}$$ Would you like to explore more or adjust these probabilities? Here are 5 questions to expand on this: 1. What are the characteristics of a discrete probability distribution? 2. How do we handle cases with multiple missing probabilities? 3. What are some other ways to distribute the remaining 0.33 between $$P(X = 0)$$ and $$P(X = 4)$$? 4. How would you verify if this is a valid probability distribution? 5. Can we use these values to compute expected value or variance? **Tip**: When assigning missing probabilities, always ensure the sum of all probabilities equals 1 and the individual values are valid (between 0 and 1).Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Distribution
Discrete Random Variables
Formulas
Sum of probabilities: P(X = -3) + P(X = 0) + P(X = 4) + P(X = 5) + P(X = 6) = 1
Theorems
Probability Distribution Properties
Suitable Grade Level
Grades 10-12
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