1. **Stating the problem:** We have measured 10 mechanical parts and recorded their deviations (gaps) from a fixed length. The gaps and their frequencies are: | Gap (mm) | 0 | 1 | 2 | 3 | |---|---|---|---|---| | Number of pieces | 1 | 4 | 3 | 2 | We need to represent this distribution in a bar diagram and calculate the mean, mode, and median. 2. **Bar diagram representation:** - The x-axis represents the gap values (0, 1, 2, 3). - The y-axis represents the number of pieces. - Bars heights correspond to frequencies: 1 for gap 0, 4 for gap 1, 3 for gap 2, and 2 for gap 3. 3. **Calculating the mean:** - The mean is the weighted average of the gaps. - Formula: $$\text{Mean} = \frac{\sum (x_i \times f_i)}{\sum f_i}$$ where $x_i$ are gap values and $f_i$ their frequencies. Calculate numerator: $$0 \times 1 + 1 \times 4 + 2 \times 3 + 3 \times 2 = 0 + 4 + 6 + 6 = 16$$ Calculate denominator: $$1 + 4 + 3 + 2 = 10$$ Mean: $$\frac{16}{10} = 1.6$$ 4. **Calculating the mode:** - The mode is the gap value with the highest frequency. - Frequencies: 1 (gap 0), 4 (gap 1), 3 (gap 2), 2 (gap 3). - Highest frequency is 4 at gap 1. Mode = 1 5. **Calculating the median:** - The median is the middle value when data is ordered. - Total pieces = 10 (even number), so median is average of 5th and 6th values. List all pieces by gap: - 1 piece at 0 - 4 pieces at 1 (positions 2 to 5) - 3 pieces at 2 (positions 6 to 8) - 2 pieces at 3 (positions 9 to 10) 5th piece gap = 1 6th piece gap = 2 Median: $$\frac{1 + 2}{2} = 1.5$$ **Final answers:** - Mean = 1.6 - Mode = 1 - Median = 1.5