1. **Problem a:** Find the midpoint of the first group 0 - 9. The midpoint of a group is the average of the lower and upper boundaries. $$\text{Midpoint} = \frac{0 + 9}{2} = \frac{9}{2} = 4.5$$ 2. **Problem b:** Calculate an estimate of the mean cost of an invoice. We use the midpoint of each group as a representative value and multiply by the frequency, then divide by total frequency. Midpoints and frequencies: - 0-9: 4.5, freq = 50 - 10-19: 14.5, freq = 204 - 20-49: 34.5, freq = 165 - 50-99: 74.5, freq = 139 - 100-149: 124.5, freq = 75 - 150-199: 174.5, freq = 62 - 200-499: 349.5, freq = 46 - 500-749: 624.5, freq = 9 Calculate total sum: $$\sum (\text{midpoint} \times \text{freq}) = 4.5 \times 50 + 14.5 \times 204 + 34.5 \times 165 + 74.5 \times 139 + 124.5 \times 75 + 174.5 \times 62 + 349.5 \times 46 + 624.5 \times 9$$ Calculate each term: - $4.5 \times 50 = 225$ - $14.5 \times 204 = 2958$ - $34.5 \times 165 = 5692.5$ - $74.5 \times 139 = 10355.5$ - $124.5 \times 75 = 9337.5$ - $174.5 \times 62 = 10819$ - $349.5 \times 46 = 16077$ - $624.5 \times 9 = 5620.5$ Sum all: $$225 + 2958 + 5692.5 + 10355.5 + 9337.5 + 10819 + 16077 + 5620.5 = 54085$$ Total frequency: $$50 + 204 + 165 + 139 + 75 + 62 + 46 + 9 = 750$$ Mean estimate: $$\frac{54085}{750} = 72.1133$$ 3. **Problem c:** Calculate an estimate for the median. Total frequency is 750, so median position is at $\frac{750}{2} = 375$th value. Cumulative frequencies: - 0-9: 50 - 10-19: 254 - 20-49: 419 The median lies in the 20-49 group because cumulative frequency reaches 419 which is above 375. Median formula for grouped data: $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times w$$ Where: - $L = 20$ (lower boundary of median group) - $N = 750$ (total frequency) - $F = 254$ (cumulative frequency before median group) - $f = 165$ (frequency of median group) - $w = 30$ (width of median group: 49 - 20 + 1 = 30) Calculate: $$\text{Median} = 20 + \left(\frac{375 - 254}{165}\right) \times 30 = 20 + \left(\frac{121}{165}\right) \times 30 = 20 + 0.7333 \times 30 = 20 + 22 = 42$$ 4. **Problem d:** Which average, mean or median best represents the data? Give a reason. The mean is approximately 72.11, the median is 42. Since the data is skewed by very large values in the higher groups (e.g., 200-499 and 500-749), the mean is pulled upward. The median better represents the central tendency because it is less affected by extreme values. --- **Summary:** - Midpoint of 0-9 group: $4.5$ - Estimated mean cost: $72.11$ - Estimated median cost: $42$ - Median best represents the data because it is less affected by skewed high values.