Question

The assets(in billions of dollars) of the four wealthiest people in a particular country are 41 ,34,21,12 Assume that samples of size n=2 are randomly selected with replacement from this population of four values

Answer 1

To compute the possible samples of size \( n=2 \) from the given population of assets (41, 34, 21, 12), we can consider the sampling process with replacement.

**Population Values:**
- \( A_1 = 41 \)
- \( A_2 = 34 \)
- \( A_3 = 21 \)
- \( A_4 = 12 \)

Since we are sampling with replacement, each sample can include any of the four values for each selection. The size of each sample is \( n=2 \), and there are \( 4 \) choices for the first pick and \( 4 \) choices for the second pick.

### Possible Samples
We need to list all combinations taking into account that the sampling is done with replacement:

1. \( (41, 41) \)
2. \( (41, 34) \)
3. \( (41, 21) \)
4. \( (41, 12) \)
5. \( (34, 41) \)
6. \( (34, 34) \)
7. \( (34, 21) \)
8. \( (34, 12) \)
9. \( (21, 41) \)
10. \( (21, 34) \)
11. \( (21, 21) \)
12. \( (21, 12) \)
13. \( (12, 41) \)
14. \( (12, 34) \)
15. \( (12, 21) \)
16. \( (12, 12) \)

### Summary
The total number of possible samples of size \( n=2 \) from the population of 4 people, when sampled with replacement, is \( 4^2 = 16 \).

These are all the combinations we could form with the given population. Each pair represents a sample drawn from the original population with replacement.