Question 1188511: The bivariate distribution of X and Y is described below:
X
Y 1 2
1 0.22 0.47
2 0.14 0.17
A. Find the marginal probability distribution of X.
1:
2:
B. Find the marginal probability distribution of Y.
1:
2:
C. Compute the mean and variance of X.
Mean =
Variance =
C. Compute the mean and variance of Y.
Mean =
Variance =
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the solution:
**A. Marginal Probability Distribution of X:**
The marginal probability distribution of X is found by summing the joint probabilities for each value of X across all values of Y.
* P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) = 0.22 + 0.47 = 0.69
* P(X=2) = P(X=2, Y=1) + P(X=2, Y=2) = 0.14 + 0.17 = 0.31
**B. Marginal Probability Distribution of Y:**
The marginal probability distribution of Y is found by summing the joint probabilities for each value of Y across all values of X.
* P(Y=1) = P(X=1, Y=1) + P(X=2, Y=1) = 0.22 + 0.14 = 0.36
* P(Y=2) = P(X=1, Y=2) + P(X=2, Y=2) = 0.47 + 0.17 = 0.64
**C. Mean and Variance of X:**
* **Mean (E[X]):** E[X] = Σ [x * P(X=x)] = (1 * 0.69) + (2 * 0.31) = 0.69 + 0.62 = 1.31
* **Variance (Var[X]):** Var[X] = E[X²] - (E[X])²
* E[X²] = Σ [x² * P(X=x)] = (1² * 0.69) + (2² * 0.31) = 0.69 + 1.24 = 1.93
* Var[X] = 1.93 - (1.31)² = 1.93 - 1.7161 = 0.2139
**D. Mean and Variance of Y:**
* **Mean (E[Y]):** E[Y] = Σ [y * P(Y=y)] = (1 * 0.36) + (2 * 0.64) = 0.36 + 1.28 = 1.64
* **Variance (Var[Y]):** Var[Y] = E[Y²] - (E[Y])²
* E[Y²] = Σ [y² * P(Y=y)] = (1² * 0.36) + (2² * 0.64) = 0.36 + 2.56 = 2.92
* Var[Y] = 2.92 - (1.64)² = 2.92 - 2.6896 = 0.2304
**Answers:**
* **A. Marginal Probability Distribution of X:**
* 1: 0.69
* 2: 0.31
* **B. Marginal Probability Distribution of Y:**
* 1: 0.36
* 2: 0.64
* **C. Mean and Variance of X:**
* Mean = 1.31
* Variance = 0.2139
* **D. Mean and Variance of Y:**
* Mean = 1.64
* Variance = 0.2304
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