1. **Problem Statement:** Calculate the average time spent to complete the exam, the range time used by the 1/4 fastest students, and the mean deviation about the mean for the given grouped frequency data of exam times for 100 students. 2. **Given Data:** | Time Interval | Frequency | |--------------|-----------| | 30-34 | 2 | | 35-39 | 4 | | 40-44 | 5 | | 45-49 | 4 | | 50-54 | 6 | | 55-59 | 9 | | 60-64 | 12 | | 65-69 | 13 | | 70-74 | 10 | | 75-79 | 12 | | 80-84 | 12 | | 85-89 | 11 | 3. **Step 1: Calculate midpoints for each class interval:** Midpoint $x_i = \frac{\text{lower limit} + \text{upper limit}}{2}$ | Interval | Midpoint $x_i$ | Frequency $f_i$ | |----------|----------------|----------------| | 30-34 | 32 | 2 | | 35-39 | 37 | 4 | | 40-44 | 42 | 5 | | 45-49 | 47 | 4 | | 50-54 | 52 | 6 | | 55-59 | 57 | 9 | | 60-64 | 62 | 12 | | 65-69 | 67 | 13 | | 70-74 | 72 | 10 | | 75-79 | 77 | 12 | | 80-84 | 82 | 12 | | 85-89 | 87 | 11 | 4. **Step 2: Calculate the mean time:** Formula for mean of grouped data: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate $\sum f_i x_i$: $$\sum f_i x_i = 2\times32 + 4\times37 + 5\times42 + 4\times47 + 6\times52 + 9\times57 + 12\times62 + 13\times67 + 10\times72 + 12\times77 + 12\times82 + 11\times87$$ Calculate each term: $64 + 148 + 210 + 188 + 312 + 513 + 744 + 871 + 720 + 924 + 984 + 957 = 6635$ Total frequency $\sum f_i = 100$ Mean: $$\bar{x} = \frac{6635}{100} = 66.35$$ 5. **Step 3: Find the range time used by the 1/4 fastest students:** 1/4 of 100 students = 25 students. We find the cumulative frequency from the smallest time upwards: | Interval | Frequency | Cumulative Frequency | |----------|-----------|----------------------| | 30-34 | 2 | 2 | | 35-39 | 4 | 6 | | 40-44 | 5 | 11 | | 45-49 | 4 | 15 | | 50-54 | 6 | 21 | | 55-59 | 9 | 30 | The 25th student lies in the 55-59 interval. The lower boundary of 55-59 is 54.5 (assuming class width 5, lower boundary = lower limit - 0.5). Number of students before this class = 21. Number of students in this class = 9. Position of 25th student in this class = 25 - 21 = 4. Using linear interpolation for the 25th student time: $$\text{Time} = 54.5 + \left(\frac{4}{9}\right) \times 5 = 54.5 + 2.22 = 56.72$$ The fastest 1/4 students took time from 30 to approximately 56.72 minutes. Range for 1/4 fastest students: $$56.72 - 30 = 26.72$$ 6. **Step 4: Calculate the mean deviation about the mean:** Mean deviation formula: $$MD = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$$ Calculate $|x_i - \bar{x}|$ and multiply by $f_i$: | $x_i$ | $f_i$ | $|x_i - 66.35|$ | $f_i \times |x_i - 66.35|$ | |-------|-------|----------------|-------------------------| | 32 | 2 | 34.35 | 68.7 | | 37 | 4 | 29.35 | 117.4 | | 42 | 5 | 24.35 | 121.75 | | 47 | 4 | 19.35 | 77.4 | | 52 | 6 | 14.35 | 86.1 | | 57 | 9 | 9.35 | 84.15 | | 62 | 12 | 4.35 | 52.2 | | 67 | 13 | 0.65 | 8.45 | | 72 | 10 | 5.65 | 56.5 | | 77 | 12 | 10.65 | 127.8 | | 82 | 12 | 15.65 | 187.8 | | 87 | 11 | 20.65 | 227.15 | Sum of $f_i |x_i - \bar{x}|$: $$68.7 + 117.4 + 121.75 + 77.4 + 86.1 + 84.15 + 52.2 + 8.45 + 56.5 + 127.8 + 187.8 + 227.15 = 1215.2$$ Mean deviation: $$MD = \frac{1215.2}{100} = 12.152$$ **Final answers:** - Average time spent: $66.35$ minutes - Range time used by 1/4 fastest students: $26.72$ minutes - Mean deviation about mean: $12.152$ minutes