https://file.aiquickdraw.com/mathbot/file/1726083110107g50lzgc0.jpg

The problem in the image involves a bivariate random variable \( (X, Y) \) with a given probability distribution. The random variable takes the following values:

- \( (1, 1) \) with probability 0.25
- \( (2, 1) \) with probability 0.25
- \( (1, 2) \) with probability 0.50

The task requires calculating:
1. The median of \( X \)
2. The variance of \( Y \)
3. The covariance of \( X \) and \( Y \)
4. The distribution function \( F(x, y) \)
5. The value of \( F(1, 2) \)
6. The distribution function of \( X \)
7. The probability function of \( Y \)

Let’s go through these calculations step-by-step:

### 1. **Median of \( X \)**:
The median is the value where half of the data lies below and half above. We can organize the distribution of \( X \):
- \( X = 1 \) with total probability \( 0.25 + 0.50 = 0.75 \)
- \( X = 2 \) with total probability \( 0.25 \)

Thus, the median is \( X = 1 \), since the cumulative probability reaches 0.75 for \( X = 1 \), exceeding 0.5.

### 2. **Variance of \( Y \)**:
We need to calculate the expected value \( E(Y) \) and then \( E(Y^2) \) to find the variance \( \text{Var}(Y) = E(Y^2) - [E(Y)]^2 \).

- \( P(Y = 1) = 0.25 + 0.25 = 0.50 \)
- \( P(Y = 2) = 0.50 \)

#### Expected value of \( Y \):
\[
E(Y) = 1(0.50) + 2(0.50) = 0.50 + 1 = 1.50
\]

#### \( E(Y^2) \):
\[
E(Y^2) = 1^2(0.50) + 2^2(0.50) = 0.50 + 4(0.50) = 0.50 + 2 = 2.50
\]

#### Variance of \( Y \):
\[
\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = 2.50 - (1.50)^2 = 2.50 - 2.25 = 0.25
\]

### 3. **Covariance of \( X \) and \( Y \)**:
Covariance is given by:
\[
\text{Cov}(X, Y) = E(XY) - E(X)E(Y)
\]

First, calculate \( E(XY) \):
\[
E(XY) = (1 \cdot 1)(0.25) + (2 \cdot 1)(0.25) + (1 \cdot 2)(0.50) = 0.25 + 0.50 + 1 = 1.75
\]

Now, calculate \( E(X) \):
\[
E(X) = 1(0.75) + 2(0.25) = 0.75 + 0.50 = 1.25
\]

Now, use the previously calculated \( E(Y) = 1.50 \).

Thus,
\[
\text{Cov}(X, Y) = 1.75 - (1.25)(1.50) = 1.75 - 1.875 = -0.125
\]

### 4. **Distribution function \( F(x, y) \)**:
The distribution function \( F(x, y) \) is the probability that \( X \leq x \) and \( Y \leq y \). We will compute it for different values of \( x \) and \( y \).

For \( F(1, 2) \), the probability that \( X \leq 1 \) and \( Y \leq 2 \):
- For \( X = 1, Y = 1 \) and \( X = 1, Y = 2 \), the combined probability is \( 0.25 + 0.50 = 0.75 \).

Thus, \( F(1, 2) = 0.75 \).

### 5. **Distribution function of \( X \)**:
The marginal distribution of \( X \) is:
- \( P(X = 1) = 0.25 + 0.50 = 0.75 \)
- \( P(X = 2) = 0.25 \)

### 6. **Probability function of \( Y \)**:
The marginal distribution of \( Y \) is:
- \( P(Y = 1) = 0.25 + 0.25 = 0.50 \)
- \( P(Y = 2) = 0.50 \)

---

Would you like any additional clarification on these solutions?

### 5 Related Questions:
1. How is the covariance between two random variables interpreted in statistics?
2. Can you explain the relationship between variance and standard deviation?
3. What is the significance of the distribution function in probability theory?
4. How would the distribution change if additional values were added for \( X \) and \( Y \)?
5. What are some common applications of bivariate distributions in real-life scenarios?

### Tip:
Covariance helps determine whether two variables increase or decrease together. If it's positive, they tend to increase together; if negative, one tends to decrease when the other increases.

The bivariate random variable (X, Y) takes the values (1, 1) with probability 0.25, (2, 1) with probability 0.25, and (1, 2) with probability 0.50. Find the median of X, the variance of Y, and the covariance of X and Y. Let F(x, y) be the distribution function of (X, Y). Find F(1, 2). Draw the distribution function of X and the probability function of Y.

Solve a bivariate random variable problem by calculating the median of X, variance of Y, and covariance between X and Y. Learn how to find the distribution function F(x, y) and understand key concepts in probability theory. Suitable for college-level statistics students.

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