1. **Calculate the mean, median, and mode for the ungrouped data.** Given data: 22, 21, 23, 24, 25, 24, 23, 22, 5, 2, 4, 3, 1, 2, 4, 16, 17, 18, 17, 18, 19, 17, 18, 16, 20, 20, 18, 10, 9, 7, 8, 6, 9, 8, 6, 9, 11, 13, 15, 12, 14, 14, 12, 13, 15, 11, 11, 14, 13, 12 - Number of data points, $n = 50$ - Mean, $\bar{x} = \frac{\sum x_i}{n}$ Calculate sum: $$\sum x_i = 22+21+23+24+25+24+23+22+5+2+4+3+1+2+4+16+17+18+17+18+19+17+18+16+20+20+18+10+9+7+8+6+9+8+6+9+11+13+15+12+14+14+12+13+15+11+11+14+13+12 = 700$$ So, $$\bar{x} = \frac{700}{50} = 14$$ - Median: Sort data and find middle value(s). Sorted data (ascending): 1,2,2,3,4,4,5,6,6,7,8,8,9,9,9,10,11,11,11,12,12,12,13,13,13,14,14,14,15,15,16,16,17,17,17,18,18,18,18,19,20,20,21,22,22,23,23,24,24,25 Since $n=50$ (even), median is average of 25th and 26th values: 25th value = 13, 26th value = 14 Median = $\frac{13+14}{2} = 13.5$ - Mode: The most frequent value(s). Frequencies: - 9 appears 3 times - 11 appears 3 times - 14 appears 3 times - 18 appears 4 times (highest) Mode = 18 2. **Construct a frequency distribution table with class width 4.** Classes (start from 0 to cover minimum 1): 0-3, 4-7, 8-11, 12-15, 16-19, 20-23, 24-27 Count frequencies: - 0-3: 1,2,2,3 → 4 - 4-7: 4,4,5,6,6,7 → 6 - 8-11: 8,8,9,9,9,10,11,11,11 → 9 - 12-15: 12,12,12,13,13,13,14,14,14,15,15 → 11 - 16-19: 16,16,17,17,17,18,18,18,18,19 → 10 - 20-23: 20,20,21,22,22,23,23 → 7 - 24-27: 24,24,25 → 3 | Class Interval | Frequency | |----------------|-----------| | 0 - 3 | 4 | | 4 - 7 | 6 | | 8 - 11 | 9 | | 12 - 15 | 11 | | 16 - 19 | 10 | | 20 - 23 | 7 | | 24 - 27 | 3 | 3. **Calculate mean, median, and mode for grouped data.** - Midpoints ($x_i$): 0-3: 1.5, 4-7: 5.5, 8-11: 9.5, 12-15: 13.5, 16-19: 17.5, 20-23: 21.5, 24-27: 25.5 - Multiply midpoints by frequencies: $1.5 \times 4 = 6$ $5.5 \times 6 = 33$ $9.5 \times 9 = 85.5$ $13.5 \times 11 = 148.5$ $17.5 \times 10 = 175$ $21.5 \times 7 = 150.5$ $25.5 \times 3 = 76.5$ Sum of $f x_i = 6 + 33 + 85.5 + 148.5 + 175 + 150.5 + 76.5 = 675$ Mean grouped: $$\bar{x} = \frac{\sum f x_i}{n} = \frac{675}{50} = 13.5$$ - Median grouped: Cumulative frequencies: 0-3: 4 4-7: 10 8-11: 19 12-15: 30 16-19: 40 20-23: 47 24-27: 50 Median class is where cumulative frequency reaches $\frac{n}{2} = 25$, which is 12-15 class. Median formula: $$\text{Median} = L + \left( \frac{\frac{n}{2} - F}{f_m} \right) \times c$$ Where: - $L = 12$ (lower boundary of median class) - $n = 50$ - $F = 19$ (cumulative frequency before median class) - $f_m = 11$ (frequency of median class) - $c = 4$ (class width) Calculate: $$\text{Median} = 12 + \left( \frac{25 - 19}{11} \right) \times 4 = 12 + \left( \frac{6}{11} \right) \times 4 = 12 + 2.18 = 14.18$$ - Mode grouped: Modal class is class with highest frequency = 12-15 (frequency 11) Mode formula: $$\text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times c$$ Where: - $L = 12$ - $f_1 = 11$ (frequency of modal class) - $f_0 = 9$ (frequency before modal class, 8-11) - $f_2 = 10$ (frequency after modal class, 16-19) - $c = 4$ Calculate: $$\text{Mode} = 12 + \left( \frac{11 - 9}{2 \times 11 - 9 - 10} \right) \times 4 = 12 + \left( \frac{2}{22 - 19} \right) \times 4 = 12 + \left( \frac{2}{3} \right) \times 4 = 12 + 2.67 = 14.67$$ 4. **Compare results:** | Statistic | Ungrouped Data | Grouped Data | |-----------|----------------|--------------| | Mean | 14 | 13.5 | | Median | 13.5 | 14.18 | | Mode | 18 | 14.67 | - The mean and median are close between grouped and ungrouped data, showing grouping smooths data slightly. - The mode differs more, as grouping affects the most frequent class representation. Grouping simplifies data but may reduce precision in mode estimation.