1. **State the problem:** We have two players' scores with given means and dot plots. We need to find: (a) The distance between the means. (b) The mean absolute deviation (MAD) for each player. (c) Express the distance between the means as a multiple of the MAD. 2. **Distance between the means:** The mean for Player A is 10 points, and for Player B is 7 points. Distance between means = $|10 - 7| = 3$ points. 3. **Calculate MAD for Player A:** Player A scores and frequencies: - 8 (4 times), 9 (1 time), 10 (2 times), 11 (1 time), 12 (4 times) Mean = 10. Calculate absolute deviations from mean: - $|8 - 10| = 2$ (4 times) - $|9 - 10| = 1$ (1 time) - $|10 - 10| = 0$ (2 times) - $|11 - 10| = 1$ (1 time) - $|12 - 10| = 2$ (4 times) Sum of absolute deviations = $2 \times 4 + 1 \times 1 + 0 \times 2 + 1 \times 1 + 2 \times 4 = 8 + 1 + 0 + 1 + 8 = 18$ Total number of scores = $4 + 1 + 2 + 1 + 4 = 12$ MAD for Player A = $\frac{18}{12} = 1.5$ points. 4. **Calculate MAD for Player B:** Player B scores and frequencies: - 5 (4 times), 6 (1 time), 7 (2 times), 8 (1 time), 9 (4 times) Mean = 7. Calculate absolute deviations from mean: - $|5 - 7| = 2$ (4 times) - $|6 - 7| = 1$ (1 time) - $|7 - 7| = 0$ (2 times) - $|8 - 7| = 1$ (1 time) - $|9 - 7| = 2$ (4 times) Sum of absolute deviations = $2 \times 4 + 1 \times 1 + 0 \times 2 + 1 \times 1 + 2 \times 4 = 8 + 1 + 0 + 1 + 8 = 18$ Total number of scores = $4 + 1 + 2 + 1 + 4 = 12$ MAD for Player B = $\frac{18}{12} = 1.5$ points. 5. **Express distance between means as multiple of MAD:** Distance between means = 3 points MAD (either player) = 1.5 points Distance between means = $\frac{3}{1.5} = 2$ times the MAD. **Final answers:** - Distance between the means: 3 points - MAD for Player A: 1.5 points - MAD for Player B: 1.5 points - Distance between the means = 2 times the MAD