Question 1191528
Here's how to analyze the given population and its samples:

**A. Listing Samples and Computing Means:**

Since the population is small, we can list all possible samples of size 2 *without replacement*.  This means a value can only appear once in each sample.

* **Samples:** (12, 12), (12, 14), (12, 16), (14, 16)
* **Sample Means:** 12, 13, 14, 15

**B. Comparing the Mean of Sample Means and the Population Mean:**

* **Population Mean:** (12 + 12 + 14 + 16) / 4 = 13.5
* **Mean of Sample Means:** (12 + 13 + 14 + 15) / 4 = 13.5

The mean of the distribution of sample means is equal to the population mean. This is a fundamental concept in statistics.

**C. Comparing Dispersion:**

* **Population Standard Deviation:**  We can calculate the population standard deviation using the following formula:
   σ = sqrt[ Σ(xi - μ)² / N ]
   Where:
     * xi are the individual values
     * μ is the population mean
     * N is the population size

   σ = sqrt[ (1.5² + 1.5² + 0.5² + 2.5²) / 4 ]
   σ = sqrt[ 2.25+2.25+0.25+6.25 /4] = sqrt(11/4) = sqrt(2.75) ≈ 1.66

* **Standard Deviation of Sample Means (Standard Error):** The standard deviation of the sample means is called the standard error. It is calculated as:
   Standard Error = σ / sqrt(n)
    Where σ is the population standard deviation and n is the sample size. 

   Standard Error = 1.66 / sqrt(2) ≈ 1.17

The dispersion (as measured by the standard deviation) of the sample means is *smaller* than the dispersion of the population.  This is also a fundamental concept: the distribution of sample means is less variable than the original population. This makes sense  -  averages tend to be more stable and less extreme than individual values.