1. **Problem Statement:** We have marks of 30 students in science: 75, 70, 55, 52, 37, 61, 42, 70, 95, 82, 45, 66, 53, 70, 47, 62, 70, 55, 85, 72, 63, 78, 60, 65, 57, 73, 87, 50, 64, 74. We need to find: - a) Mode of the data. - b) Median using a frequency table with class intervals of 10. - c) Frequency polygon (not plotted here, but frequency data prepared). --- 2. **Finding the Mode:** - Mode is the value that appears most frequently. - Count occurrences of each mark: - 70 appears 5 times (most frequent). **Answer:** Mode = $70$ --- 3. **Constructing Frequency Table with Class Intervals of 10:** - Class intervals: 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99 - Count frequencies: - 30-39: 1 (37) - 40-49: 3 (42, 45, 47) - 50-59: 5 (50, 52, 53, 55, 55, 57) actually 6 - 60-69: 7 (60, 61, 62, 63, 64, 65, 66) - 70-79: 7 (70, 70, 70, 70, 72, 73, 74, 75, 78) actually 9 - 80-89: 3 (82, 85, 87) - 90-99: 1 (95) Correct frequencies: - 30-39: 1 - 40-49: 3 - 50-59: 6 - 60-69: 7 - 70-79: 9 - 80-89: 3 - 90-99: 1 --- 4. **Calculate Median:** - Total data points $n=30$ - Median class is where cumulative frequency reaches or exceeds $\frac{n}{2} = 15$ Cumulative frequencies: - 30-39: 1 - 40-49: 4 - 50-59: 10 - 60-69: 17 (median class) Median class: 60-69 Use median formula: $$\text{Median} = L + \left(\frac{\frac{n}{2} - F}{f}\right) \times h$$ Where: - $L=60$ (lower boundary of median class) - $n=30$ - $F=10$ (cumulative frequency before median class) - $f=7$ (frequency of median class) - $h=10$ (class width) Calculate: $$\text{Median} = 60 + \left(\frac{15 - 10}{7}\right) \times 10 = 60 + \frac{5}{7} \times 10 = 60 + 7.14 = 67.14$$ **Answer:** Median $\approx 67.14$ --- 5. **Frequency Polygon:** - Plot points at class midpoints vs frequency: - Midpoints: 34.5, 44.5, 54.5, 64.5, 74.5, 84.5, 94.5 - Frequencies: 1, 3, 6, 7, 9, 3, 1 This data can be used to draw the frequency polygon by joining these points with straight lines. --- **Summary:** - Mode = $70$ - Median $\approx 67.14$ - Frequency polygon points prepared for plotting.