1. **Problem Statement:** We have data on hours spent studying by 40 students: 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 2, 2, 3, 4, 4, 5, 6, 8, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. We will: a) Construct a frequency distribution table. b) Add cumulative frequency. c) Compute mean, median, and mode. 2. **Frequency Distribution Table:** Count each unique hour value: | Hours (x) | Frequency (f) | |-----------|---------------| | 1 | 2 | | 2 | 5 | | 3 | 5 | | 4 | 6 | | 5 | 5 | | 6 | 6 | | 7 | 4 | | 8 | 4 | | 9 | 2 | | 10 | 1 | 3. **Cumulative Frequency (CF):** Add frequencies cumulatively: | Hours (x) | Frequency (f) | Cumulative Frequency (CF) | |-----------|---------------|---------------------------| | 1 | 2 | 2 | | 2 | 5 | 7 | | 3 | 5 | 12 | | 4 | 6 | 18 | | 5 | 5 | 23 | | 6 | 6 | 29 | | 7 | 4 | 33 | | 8 | 4 | 37 | | 9 | 2 | 39 | | 10 | 1 | 40 | 4. **Mean Calculation:** Formula: $$\text{Mean} = \frac{\sum f x}{\sum f}$$ Calculate $$\sum f x$$: $$2\times1 + 5\times2 + 5\times3 + 6\times4 + 5\times5 + 6\times6 + 4\times7 + 4\times8 + 2\times9 + 1\times10 = 2 + 10 + 15 + 24 + 25 + 36 + 28 + 32 + 18 + 10 = 200$$ Total frequency $$\sum f = 40$$ Mean: $$\frac{200}{40} = 5$$ 5. **Median Calculation:** Median position: $$\frac{n+1}{2} = \frac{40+1}{2} = 20.5$$th value From CF, 20.5 lies in the class where CF just exceeds 20.5, which is at hours = 5 (CF=23). So, median = 5 hours. 6. **Mode Calculation:** Mode is the value with highest frequency. Highest frequency is 6, which occurs at hours 4 and 6. So, the data is bimodal with modes 4 and 6. **Final answers:** - Frequency distribution and cumulative frequency as above. - Mean = 5 - Median = 5 - Mode = 4 and 6 (bimodal)