1. **Problem Statement:** The school wants to analyze learners' performance based on the histogram data showing learners' scores and frequencies. 2. **Given Data:** Class intervals (scores): 24.5-34.5, 34.5-44.5, 44.5-54.5, 54.5-64.5, 64.5-74.5, 74.5-84.5, 84.5-94.5 Frequencies (approximate): 13, 29, 12, 45, 85, 29, 12 3. **Step 1: Calculate the midpoint for each class interval** Midpoint formula: $$\text{Midpoint} = \frac{\text{Lower boundary} + \text{Upper boundary}}{2}$$ - 24.5 to 34.5: $\frac{24.5 + 34.5}{2} = 29.5$ - 34.5 to 44.5: $\frac{34.5 + 44.5}{2} = 39.5$ - 44.5 to 54.5: $\frac{44.5 + 54.5}{2} = 49.5$ - 54.5 to 64.5: $\frac{54.5 + 64.5}{2} = 59.5$ - 64.5 to 74.5: $\frac{64.5 + 74.5}{2} = 69.5$ - 74.5 to 84.5: $\frac{74.5 + 84.5}{2} = 79.5$ - 84.5 to 94.5: $\frac{84.5 + 94.5}{2} = 89.5$ 4. **Step 2: Calculate the total number of learners** $$N = 13 + 29 + 12 + 45 + 85 + 29 + 12 = 225$$ 5. **Step 3: Calculate the mean score** Mean formula for grouped data: $$\bar{x} = \frac{\sum f_i x_i}{N}$$ where $f_i$ is frequency and $x_i$ is midpoint. Calculate $\sum f_i x_i$: $$13 \times 29.5 = 383.5$$ $$29 \times 39.5 = 1145.5$$ $$12 \times 49.5 = 594$$ $$45 \times 59.5 = 2677.5$$ $$85 \times 69.5 = 5907.5$$ $$29 \times 79.5 = 2305.5$$ $$12 \times 89.5 = 1074$$ Sum: $$383.5 + 1145.5 + 594 + 2677.5 + 5907.5 + 2305.5 + 1074 = 13987.5$$ Mean: $$\bar{x} = \frac{13987.5}{225} \approx 62.17$$ 6. **Step 4: Calculate the percentage of learners below the pass mark 55** Pass mark 55 lies between 54.5 and 64.5 intervals. Frequencies below 55 are for intervals: - 24.5-34.5: 13 - 34.5-44.5: 29 - 44.5-54.5: 12 Total learners below 55: $$13 + 29 + 12 = 54$$ Percentage below pass mark: $$\frac{54}{225} \times 100 \approx 24\%$$ 7. **Interpretation:** - The mean score is approximately 62.17. - About 24% of learners scored below the pass mark of 55. If the previous mean score was lower than 62.17, this indicates an improvement in learners' performance. **Final answers:** - Mean score: $62.17$ - Percentage below pass mark: $24\%$