1. **State the problem:** We are given a frequency distribution table with cumulative frequencies for 60 scores. We need to find the missing frequency value represented by '?'. 2. **Understand the data:** The cumulative frequency (Cumulative f) is the running total of frequencies (F) up to that class interval. 3. **Use the cumulative frequency property:** The cumulative frequency for the class interval 55-59 is 7, and the cumulative frequency for the previous class interval 50-54 is 3. 4. **Calculate the missing frequency:** The frequency for 55-59 is the difference between the cumulative frequency at 55-59 and the cumulative frequency at 50-54. $$F_{55-59} = Cumulative\ f_{55-59} - Cumulative\ f_{50-54}$$ $$F_{55-59} = 7 - 3 = 4$$ 5. **Answer:** The missing frequency value '?' is 4. --- 1. **State the problem:** Find the P10 (10th percentile) value for the frequency distribution. 2. **Recall the formula for percentile position:** $$L = \text{lower boundary of the class containing the percentile}$$ $$N = \text{total number of observations} = 60$$ $$P = 10\% = 0.10$$ $$CF = \text{cumulative frequency before the percentile class}$$ $$f = \text{frequency of the percentile class}$$ $$w = \text{class width}$$ Percentile formula: $$P_{10} = L + \left(\frac{PN - CF}{f}\right) \times w$$ 3. **Find the class containing P10:** Calculate $PN = 0.10 \times 60 = 6$. Look at cumulative frequencies to find the class where cumulative frequency just exceeds 6. Cumulative frequencies: - 45-49: 1 - 50-54: 3 - 55-59: 7 Since 7 is the first cumulative frequency greater than or equal to 6, the P10 lies in class 55-59. 4. **Identify values:** - $L = 54.5$ (lower boundary of 55-59 class) - $CF = 3$ (cumulative frequency before 55-59) - $f = 4$ (frequency of 55-59 class) - $w = 5$ (class width) 5. **Calculate P10:** $$P_{10} = 54.5 + \left(\frac{6 - 3}{4}\right) \times 5 = 54.5 + \left(\frac{3}{4}\right) \times 5 = 54.5 + 3.75 = 58.25$$ 6. **Answer:** The P10 value is 58.25.