1. **State the problem:** We have a frequency table of weights of 71 tortoises and their frequencies. We add one tortoise weighing 39 pounds to create a new data set of 72 tortoises. We need to compare the mean and median of the new data set to the original. 2. **Calculate the original mean:** The mean is given by $$\text{mean} = \frac{\sum (\text{weight} \times \text{frequency})}{\text{total frequency}}$$ Calculate the sum of weights times frequencies: $$\sum (w \times f) = 13\times12 + 14\times8 + 15\times5 + 16\times7 + 17\times9 + 18\times10 + 19\times13 + 20\times7$$ Calculate each term: $$156 + 112 + 75 + 112 + 153 + 180 + 247 + 140 = 1175$$ Total frequency = 71 So original mean: $$\text{mean}_\text{original} = \frac{1175}{71} \approx 16.55$$ 3. **Calculate the new mean:** Add the new weight 39: $$\text{sum new} = 1175 + 39 = 1214$$ New total frequency = 72 New mean: $$\text{mean}_\text{new} = \frac{1214}{72} \approx 16.86$$ Since 16.86 > 16.55, the mean increases. 4. **Calculate the original median:** The median is the middle value when data is ordered. Total frequency is 71, so median is the 36th value. Cumulative frequencies: - Up to 13: 12 - Up to 14: 12 + 8 = 20 - Up to 15: 20 + 5 = 25 - Up to 16: 25 + 7 = 32 - Up to 17: 32 + 9 = 41 The 36th value lies in the 17-pound group (since 32 < 36 ≤ 41), so median is 17. 5. **Calculate the new median:** Now total frequency is 72, so median is average of 36th and 37th values. Cumulative frequencies remain the same up to 17 pounds (41), so both 36th and 37th values are still 17. Therefore, median remains 17. 6. **Conclusion:** The mean increases, the median stays the same. **Answer:** B. The mean of the new data set is greater than the mean of the original data set, and the medians of the two data sets are equal.