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To solve this problem, let's follow these steps:

### Step 1: Identify Possible Samples
The assets of the four wealthiest individuals are: 36, 28, 22, and 12 (in billions). Since samples of size \( n = 2 \) are selected **with replacement**, each pair combination can be drawn independently. This yields \( 4 \times 4 = 16 \) possible samples.

### Step 2: Calculate Sample Means
Each possible pair will have an average, computed by \((x_1 + x_2) / 2\), where \( x_1 \) and \( x_2 \) are the chosen values from the set \{36, 28, 22, 12\}.

#### Possible Pairs and Means:
1. (36, 36) → \( \frac{36 + 36}{2} = 36 \)
2. (36, 28) → \( \frac{36 + 28}{2} = 32 \)
3. (36, 22) → \( \frac{36 + 22}{2} = 29 \)
4. (36, 12) → \( \frac{36 + 12}{2} = 24 \)
5. (28, 36) → \( \frac{28 + 36}{2} = 32 \)
6. (28, 28) → \( \frac{28 + 28}{2} = 28 \)
7. (28, 22) → \( \frac{28 + 22}{2} = 25 \)
8. (28, 12) → \( \frac{28 + 12}{2} = 20 \)
9. (22, 36) → \( \frac{22 + 36}{2} = 29 \)
10. (22, 28) → \( \frac{22 + 28}{2} = 25 \)
11. (22, 22) → \( \frac{22 + 22}{2} = 22 \)
12. (22, 12) → \( \frac{22 + 12}{2} = 17 \)
13. (12, 36) → \( \frac{12 + 36}{2} = 24 \)
14. (12, 28) → \( \frac{12 + 28}{2} = 20 \)
15. (12, 22) → \( \frac{12 + 22}{2} = 17 \)
16. (12, 12) → \( \frac{12 + 12}{2} = 12 \)

### Step 3: Determine Frequency and Probability
Let's count how many times each unique mean occurs and then calculate its probability as a fraction of the total samples (16).

- **36** appears 1 time → Probability \( = \frac{1}{16} \)
- **32** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **29** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **28** appears 1 time → Probability \( = \frac{1}{16} \)
- **25** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **24** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **22** appears 1 time → Probability \( = \frac{1}{16} \)
- **20** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **17** appears 2 times → Probability \( = \frac{2}{16} = \frac{1}{8} \)
- **12** appears 1 time → Probability \( = \frac{1}{16} \)

### Step 4: Complete the Table

| \( \bar{x} \) | Probability |
|---------------|-------------|
| 36            | \( \frac{1}{16} \) |
| 32            | \( \frac{1}{8} \) |
| 29            | \( \frac{1}{8} \) |
| 28            | \( \frac{1}{16} \) |
| 25            | \( \frac{1}{8} \) |
| 24            | \( \frac{1}{8} \) |
| 22            | \( \frac{1}{16} \) |
| 20            | \( \frac{1}{8} \) |
| 17            | \( \frac{1}{8} \) |
| 12            | \( \frac{1}{16} \) |

Would you like further details on any part of this solution, or have any additional questions?

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Here are some related questions that might deepen your understanding:

1. How would the probabilities change if the sample was taken **without replacement**?
2. How can we use the concept of **expected value** with this sampling distribution?
3. How would this distribution change if there were more than four people in the population?
4. What is the **variance** of this sampling distribution?
5. How does the **central limit theorem** apply to larger sample sizes in this context?

**Tip:** When calculating probabilities for a sampling distribution, always check if the samples are with or without replacement as it significantly affects outcomes.

The assets (in billions of dollars) of the four wealthiest people in a particular country are 36, 28, 22, 12. Assume that samples of size n = 2 are randomly selected with replacement from this population of four values. Construct a table representing the sampling distribution of the sample mean by identifying the 16 different possible samples and finding the mean of each sample.

Learn how to create a sampling distribution of the sample mean for a population of asset values (36, 28, 22, 12) with replacement. This exercise demonstrates statistical concepts such as calculating sample means, probability distribution, and the Law of Large Numbers, suitable for high school students in Grades 10-12.

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