A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let *X* denote the number of hoses being used on the self-service island at a particular time, and let *Y* denote the number of hoses on the full-service island in use at that time. The joint pmf of *X* and *Y* appears in the accompanying tabulation.

*y*

*p*(*x*, *y*)

0

1

2

*x*

0

0.10

0.03

0.01

1

0.06

0.20

0.07

2

0.06

0.14

0.33

Given the joint probability mass function (pmf) of \(X\) (the number of hoses being used at the self-service island) and \(Y\) (the number of hoses being used at the full-service island), the table looks like this:

| \(*x*, *y*\)  | \(Y = 0\) | \(Y = 1\) | \(Y = 2\) |
|:-------------:|:---------:|:---------:|:---------:|
| \(X = 0\)     | 0.10      | 0.03      | 0.01      |
| \(X = 1\)     | 0.06      | 0.20      | 0.07      |
| \(X = 2\)     | 0.06      | 0.14      | 0.33      |

### 1. **Check for Validity of pmf:**

To confirm if this is a valid pmf, the sum of all joint probabilities must equal 1:
\[
\sum_{x=0}^{2} \sum_{y=0}^{2} p(x, y) = 1
\]
Let’s sum the values in the table:
\[
0.10 + 0.03 + 0.01 + 0.06 + 0.20 + 0.07 + 0.06 + 0.14 + 0.33 = 1.00
\]
Thus, the pmf is valid.

### 2. **Marginal pmf of \(X\):**

The marginal pmf of \(X\) is found by summing over all values of \(Y\):
\[
p_X(x) = \sum_{y} p(x, y)
\]
- \(p_X(0) = 0.10 + 0.03 + 0.01 = 0.14\)
- \(p_X(1) = 0.06 + 0.20 + 0.07 = 0.33\)
- \(p_X(2) = 0.06 + 0.14 + 0.33 = 0.53\)

So, the marginal pmf of \(X\) is:
\[
p_X(0) = 0.14, \quad p_X(1) = 0.33, \quad p_X(2) = 0.53
\]

### 3. **Marginal pmf of \(Y\):**

Similarly, the marginal pmf of \(Y\) is found by summing over all values of \(X\):
\[
p_Y(y) = \sum_{x} p(x, y)
\]
- \(p_Y(0) = 0.10 + 0.06 + 0.06 = 0.22\)
- \(p_Y(1) = 0.03 + 0.20 + 0.14 = 0.37\)
- \(p_Y(2) = 0.01 + 0.07 + 0.33 = 0.41\)

So, the marginal pmf of \(Y\) is:
\[
p_Y(0) = 0.22, \quad p_Y(1) = 0.37, \quad p_Y(2) = 0.41
\]

### 4. **Expected Values \(E[X]\) and \(E[Y]\):**

The expected value \(E[X]\) is given by:
\[
E[X] = \sum_{x} x \cdot p_X(x)
\]
\[
E[X] = 0 \cdot 0.14 + 1 \cdot 0.33 + 2 \cdot 0.53 = 0 + 0.33 + 1.06 = 1.39
\]

The expected value \(E[Y]\) is given by:
\[
E[Y] = \sum_{y} y \cdot p_Y(y)
\]
\[
E[Y] = 0 \cdot 0.22 + 1 \cdot 0.37 + 2 \cdot 0.41 = 0 + 0.37 + 0.82 = 1.19
\]

### 5. **Covariance \(\text{Cov}(X, Y)\):**

To calculate \(\text{Cov}(X, Y)\), we need \(E[XY]\), which is computed as:
\[
E[XY] = \sum_{x} \sum_{y} x \cdot y \cdot p(x, y)
\]
Let’s compute \(E[XY]\) based on the pmf table:
\[
E[XY] = (0 \cdot 0 \cdot 0.10) + (0 \cdot 1 \cdot 0.03) + (0 \cdot 2 \cdot 0.01) + (1 \cdot 0 \cdot 0.06) + (1 \cdot 1 \cdot 0.20) + (1 \cdot 2 \cdot 0.07) + (2 \cdot 0 \cdot 0.06) + (2 \cdot 1 \cdot 0.14) + (2 \cdot 2 \cdot 0.33)
\]
\[
E[XY] = 0 + 0 + 0 + 0 + 0.20 + 0.14 + 0 + 0.28 + 1.32 = 1.94
\]

Now, use the formula for covariance:
\[
\text{Cov}(X, Y) = E[XY] - E[X]E[Y]
\]
\[
\text{Cov}(X, Y) = 1.94 - (1.39 \cdot 1.19) = 1.94 - 1.6541 = 0.2859
\]

### Recap:
- The marginal pmf of \(X\) is \(p_X(0) = 0.14\), \(p_X(1) = 0.33\), \(p_X(2) = 0.53\).
- The marginal pmf of \(Y\) is \(p_Y(0) = 0.22\), \(p_Y(1) = 0.37\), \(p_Y(2) = 0.41\).
- \(E[X] = 1.39\) and \(E[Y] = 1.19\).
- \(\text{Cov}(X, Y) = 0.2859\).

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the correlation coefficient between \(X\) and \(Y\)?
2. How would you calculate \(P(X = 1 \text{ and } Y = 2)\)?
3. Can \(X\) and \(Y\) be considered independent based on the joint pmf?
4. What is the variance of \(X\) and \(Y\)?
5. How would you calculate \(P(X \leq 1, Y \geq 1)\)?

**Tip**: Covariance only measures the direction of the relationship between variables, while the correlation coefficient normalizes this to a range of \(-1\) to \(1\).

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation.

Explore the joint probability distribution for the number of hoses in use on self-service and full-service islands. This problem involves calculating marginal pmfs, expected values, and covariance based on the provided joint pmf. Aimed at undergraduate students studying statistics or probability.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation