1. **Stating the problem:** We have a dataset of 40 salesmen's undergraduate units earned: 48, 60, 96, 78, 120, 54, 72, 72, 102, 108, 96, 107, 52, 60, 66, 120, 41, 60, 118, 65, 85, 54, 125, 84, 93, 66, 48, 83, 84, 90, 79, 86, 93, 110, 120, 86, 99, 88, 73, 88. We need to: - Construct a **frequency distribution**, - Find the **mean**. 2. **Constructing a frequency distribution:** Let's choose class intervals with width 10 starting from 40 (lowest data point is 41): - 40-49 - 50-59 - 60-69 - 70-79 - 80-89 - 90-99 - 100-109 - 110-119 - 120-129 Counting data points: - 40-49: 48,48,41 → 3 - 50-59: 54,54,52 → 3 - 60-69: 60,60,60,66,66,65 → 6 - 70-79: 72,72,78,73,79 → 5 - 80-89: 85,83,84,84,86,86,88,88 → 8 - 90-99: 93,93,96,96,99,90 → 6 - 100-109: 102,107,108 → 3 - 110-119: 110,118 → 2 - 120-129: 120,120,120,125 → 4 Frequency distribution table: | Class Interval | Frequency | |---------------|-----------| | 40 - 49 | 3 | | 50 - 59 | 3 | | 60 - 69 | 6 | | 70 - 79 | 5 | | 80 - 89 | 8 | | 90 - 99 | 6 | | 100 - 109 | 3 | | 110 - 119 | 2 | | 120 - 129 | 4 | 3. **Finding the mean:** Mean formula for grouped data: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Where $f_i$ = frequency, $x_i$ = class midpoint. Class midpoints: - 40-49 → 44.5 - 50-59 → 54.5 - 60-69 → 64.5 - 70-79 → 74.5 - 80-89 → 84.5 - 90-99 → 94.5 - 100-109 → 104.5 - 110-119 → 114.5 - 120-129 → 124.5 Calculate $f_i x_i$: - 3 * 44.5 = 133.5 - 3 * 54.5 = 163.5 - 6 * 64.5 = 387 - 5 * 74.5 = 372.5 - 8 * 84.5 = 676 - 6 * 94.5 = 567 - 3 * 104.5 = 313.5 - 2 * 114.5 = 229 - 4 * 124.5 = 498 Sum frequencies $\sum f_i = 40$ Sum $f_i x_i = 133.5 + 163.5 + 387 + 372.5 + 676 + 567 + 313.5 + 229 + 498 = 3340$ (approximately) Calculate mean: $$\bar{x} = \frac{3340}{40} = 83.5$$ **Final answer:** The mean number of undergraduate units earned is approximately $83.5$.