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Number of days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Answer :
Class interval | Frequency \( (f_i) \) | Mid-point \( (x_i) \) | \( f_i x_i \) |
0-6 | 11 | 3 | 33 |
6-10 | 10 | 8 | 80 |
10-14 | 7 | 12 | 84 |
14-20 | 4 | 17 | 68 |
20-28 | 4 | 24 | 96 |
28-38 | 3 | 33 | 99 |
38-40 | 1 | 39 | 39 |
\( \sum f_i \ = \ 40 \) | \( \sum f_i x_i \ = \ 499 \) |
The mean formula is,
\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)
\( = \ \frac{499}{40} \)
\(= \ 12.48 \) days
\(\therefore \) The mean number of days a student was absent = 12.48.