1. **Problem Statement:** We have data on the amount of money 100 students spent on cellphone data in a week, grouped into class intervals with frequencies and midpoints. We need to find: (i) a) The modal class. b) An estimate of the mean amount spent. (ii) Explain why a reported median of 84 could be correct. 2. **Modal Class:** The modal class is the class interval with the highest frequency. From the table, frequencies are: 7, 11, 31, 29, 22. The highest frequency is 31, corresponding to the class 70 < x ≤ 80. So, the modal class is **70 < x ≤ 80**. 3. **Mean Estimate:** The formula for the mean from grouped data is: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is the frequency and $x_i$ is the midpoint of each class. Calculate $\sum f_i x_i$: $7 \times 55 = 385$ $11 \times 65 = 715$ $31 \times 75 = 2325$ $29 \times 85 = 2465$ $22 \times 95 = 2090$ Sum these: $$385 + 715 + 2325 + 2465 + 2090 = 7980$$ Total number of students $\sum f_i = 100$. Calculate mean: $$\bar{x} = \frac{7980}{100} = 79.80$$ So, the estimated mean amount spent is **79.80**. 4. **Median Explanation:** The median is the middle value when data is ordered. Since the total frequency is 100, the median is the average of the 50th and 51st values. Looking at cumulative frequencies: - Up to 60: 7 - Up to 70: 7 + 11 = 18 - Up to 80: 18 + 31 = 49 - Up to 90: 49 + 29 = 78 The 50th and 51st values lie in the class 80 < x ≤ 90. This class ranges from 80 to 90, so the median must be between 80 and 90. Damion's reported median of 84 lies within this interval, so it could be correct. **Final answers:** (i) a) Modal class: 70 < x ≤ 80 b) Estimated mean: 79.80 (ii) Median of 84 is plausible because it lies in the median class 80 < x ≤ 90.