1. **State the problem:** We have a population with numbers 8, 12, 7, 13, 9, and 25. We want to list all possible samples of size 3 and compute the mean of each sample. Then, construct a frequency distribution of these means. 2. **Number of possible samples:** The number of ways to choose 3 elements from 6 is given by the combination formula: $$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ 3. **List all samples and their means:** Samples and their means are: - (8,12,7): mean = $\frac{8+12+7}{3} = \frac{27}{3} = 9$ - (8,12,13): mean = $\frac{8+12+13}{3} = \frac{33}{3} = 11$ - (8,12,9): mean = $\frac{8+12+9}{3} = \frac{29}{3} \approx 9.67$ - (8,12,25): mean = $\frac{8+12+25}{3} = \frac{45}{3} = 15$ - (8,7,13): mean = $\frac{8+7+13}{3} = \frac{28}{3} \approx 9.33$ - (8,7,9): mean = $\frac{8+7+9}{3} = \frac{24}{3} = 8$ - (8,7,25): mean = $\frac{8+7+25}{3} = \frac{40}{3} \approx 13.33$ - (8,13,9): mean = $\frac{8+13+9}{3} = \frac{30}{3} = 10$ - (8,13,25): mean = $\frac{8+13+25}{3} = \frac{46}{3} \approx 15.33$ - (8,9,25): mean = $\frac{8+9+25}{3} = \frac{42}{3} = 14$ - (12,7,13): mean = $\frac{12+7+13}{3} = \frac{32}{3} \approx 10.67$ - (12,7,9): mean = $\frac{12+7+9}{3} = \frac{28}{3} \approx 9.33$ - (12,7,25): mean = $\frac{12+7+25}{3} = \frac{44}{3} \approx 14.67$ - (12,13,9): mean = $\frac{12+13+9}{3} = \frac{34}{3} \approx 11.33$ - (12,13,25): mean = $\frac{12+13+25}{3} = \frac{50}{3} \approx 16.67$ - (12,9,25): mean = $\frac{12+9+25}{3} = \frac{46}{3} \approx 15.33$ - (7,13,9): mean = $\frac{7+13+9}{3} = \frac{29}{3} \approx 9.67$ - (7,13,25): mean = $\frac{7+13+25}{3} = \frac{45}{3} = 15$ - (7,9,25): mean = $\frac{7+9+25}{3} = \frac{41}{3} \approx 13.67$ - (13,9,25): mean = $\frac{13+9+25}{3} = \frac{47}{3} \approx 15.67$ 4. **Construct frequency distribution:** Group means and count their frequencies: - 8: 1 - 9: 1 - 9.33: 2 - 9.67: 2 - 10: 1 - 10.67: 1 - 11: 1 - 11.33: 1 - 13.33: 1 - 13.67: 1 - 14: 1 - 14.67: 1 - 15: 2 - 15.33: 2 - 15.67: 1 - 16.67: 1 This frequency distribution shows how often each sample mean occurs among all possible samples of size 3.