1. **State the problem:** We have a population with values $\{2,4,8,10,5\}$. We want to list all possible samples of size 3, find their sample means, and construct the sampling distribution and probability distribution of these sample means. 2. **List all possible samples of size 3:** Since the population size is 5, the number of samples of size 3 without replacement is $\binom{5}{3} = 10$. The samples are: $\{2,4,8\}, \{2,4,10\}, \{2,4,5\}, \{2,8,10\}, \{2,8,5\}, \{2,10,5\}, \{4,8,10\}, \{4,8,5\}, \{4,10,5\}, \{8,10,5\}$. 3. **Calculate the sample means:** - $\bar{x}_1 = \frac{2+4+8}{3} = \frac{14}{3} = 4.67$ - $\bar{x}_2 = \frac{2+4+10}{3} = \frac{16}{3} = 5.33$ - $\bar{x}_3 = \frac{2+4+5}{3} = \frac{11}{3} = 3.67$ - $\bar{x}_4 = \frac{2+8+10}{3} = \frac{20}{3} = 6.67$ - $\bar{x}_5 = \frac{2+8+5}{3} = \frac{15}{3} = 5$ - $\bar{x}_6 = \frac{2+10+5}{3} = \frac{17}{3} = 5.67$ - $\bar{x}_7 = \frac{4+8+10}{3} = \frac{22}{3} = 7.33$ - $\bar{x}_8 = \frac{4+8+5}{3} = \frac{17}{3} = 5.67$ - $\bar{x}_9 = \frac{4+10+5}{3} = \frac{19}{3} = 6.33$ - $\bar{x}_{10} = \frac{8+10+5}{3} = \frac{23}{3} = 7.67$ 4. **Construct the sampling distribution of the sample mean:** The sample means are $\{3.67,4.67,5,5.33,5.67,5.67,6.33,6.67,7.33,7.67\}$. 5. **Construct the probability distribution of the sample means:** Each sample is equally likely with probability $\frac{1}{10}$. Grouping identical means: - $3.67$ with probability $\frac{1}{10}$ - $4.67$ with probability $\frac{1}{10}$ - $5$ with probability $\frac{1}{10}$ - $5.33$ with probability $\frac{1}{10}$ - $5.67$ appears twice, so probability $\frac{2}{10} = 0.2$ - $6.33$ with probability $\frac{1}{10}$ - $6.67$ with probability $\frac{1}{10}$ - $7.33$ with probability $\frac{1}{10}$ - $7.67$ with probability $\frac{1}{10}$ This completes the sampling distribution and probability distribution of the sample means.