1. **Problem Statement:** We have two frequency distributions: one for scores of candidates and another for tyre life versus type. 2. **Scores Distribution:** Scores intervals: 0-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99 Number of candidates: 40, 60, 80, 100, 64, 32, 16, 8 3. **Tyre Life Distribution:** Life intervals (in thousands km): 4-8, 8-12, 12-16, 16-20, 20-24 Type A counts: 27, 33, 39, 38, 13 Type B counts: 22, 36, 45, 27, 20 4. **Understanding the Data:** These are grouped frequency tables. To analyze or graph, we often use midpoints of intervals as representative values. 5. **Calculate Midpoints:** For scores: $$\text{Midpoints} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$ So midpoints are: $$14.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5, 94.5$$ For tyre life: $$6, 10, 14, 18, 22$$ 6. **Example Computation: Mean Score** Formula for mean of grouped data: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is frequency and $x_i$ is midpoint. Calculate numerator: $$40 \times 14.5 + 60 \times 34.5 + 80 \times 44.5 + 100 \times 54.5 + 64 \times 64.5 + 32 \times 74.5 + 16 \times 84.5 + 8 \times 94.5$$ $$= 580 + 2070 + 3560 + 5450 + 4128 + 2384 + 1352 + 756 = 22280$$ Sum of frequencies: $$40 + 60 + 80 + 100 + 64 + 32 + 16 + 8 = 400$$ Mean score: $$\bar{x} = \frac{22280}{400} = 55.7$$ 7. **Example Computation: Total Tyre Counts** Sum Type A: $$27 + 33 + 39 + 38 + 13 = 150$$ Sum Type B: $$22 + 36 + 45 + 27 + 20 = 150$$ Total tyres tested = 150 + 150 = 300 8. **Summary:** - Mean score of candidates is approximately $55.7$. - Total tyres tested are $300$, equally split between Type A and Type B. You can use these midpoints and frequencies to plot histograms or calculate other statistics like median or mode if needed.