1. **Problem statement:** We have distances travelled to school by 26 students recorded in intervals. We need to complete the frequency table, draw a histogram, calculate a probability, and decide the best average for estimating transportation cost. 2. **Complete the frequency table:** Count how many distances fall into each interval. - 1 - 5 km: 1 (given) - 6 - 10 km: 2 (given) - 11 - 15 km: 4 (given) - 16 - 20 km: 6 (given) - 21 - 25 km: Count distances 21, 22, 22, 22, 23, 24, 25 → 7 students - 26 - 30 km: Count distances 26, 28, 30 → 3 students - 31 - 35 km: Count distances 32, 34 → 2 students - 36 - 40 km: Count distance 39 → 1 student 3. **Histogram scales:** Horizontal axis: 2 cm represents 5 km. Vertical axis: 1 cm represents 1 student. Draw bars for each interval with height equal to frequency in cm. 4. **Calculate probability of distance 26 km or more:** Number of students with distance $\geq 26$ km = frequencies in intervals 26-30, 31-35, 36-40 = $3 + 2 + 1 = 6$ Total students = 26 Probability = $\frac{6}{26} = \frac{3}{13} \approx 0.2308$ 5. **Best average for estimating cost:** The mode is most appropriate because it represents the most common distance travelled, which helps estimate typical transportation cost. Mean can be skewed by extreme values, median gives middle value but mode reflects the most frequent distance. **Final answers:** (a) Frequency table completed: | Distance in kilometres | Frequency | |-----------------------|-----------| | 1 - 5 | 1 | | 6 - 10 | 2 | | 11 - 15 | 4 | | 16 - 20 | 6 | | 21 - 25 | 7 | | 26 - 30 | 3 | | 31 - 35 | 2 | | 36 - 40 | 1 | (c) Probability = $\frac{6}{26} = \frac{3}{13} \approx 0.2308$ (d) Mode is most appropriate for estimating cost because it reflects the most common distance travelled.