Question of 20
Q.Following table gives the marks obtained by 100 students in statistics subject.
Marks : 20 10 50 40 30
No. of students : 10 5 25 20 40
Find the average and median marks of students.
Or
Compute mean deviation from mean of the following data :
Marks : 0–10 10–20 20–30 30–40 40–50
No. of students : 5 10 20 5 10
Manipur CohsemCOHSEM Manipur Higher Secondary 1st Year (Commerce) 2020Subjective· 8mImportance★★★★★est
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Start your 14-day free trial to unlock the full solution →Sorting the discrete series by marks and using the direct method: Mean = Σfx/Σf = 3500/100 = 35; Median (the value where cumulative frequency first reaches N/2 = 50) = 30.
Step 1 — Arrange the series by marks (ascending) with cumulative frequency (cf):
| Marks (x) | No. of students (f) | fx | cf |
|---|---|---|---|
| 10 | 5 | 50 | 5 |
| 20 | 10 | 200 | 15 |
| 30 | 40 | 1200 | 55 |
| 40 | 20 | 800 | 75 |
| 50 | 25 | 1250 | 100 |
| Total | Σf = N = 100 | Σfx = 3500 |
Average (Arithmetic Mean):
Mean (x̄) = Σfx / Σf = 3500 / 100 = 35 marks
Median (discrete series): Locate N/2 = 100/2 = 50th item using the cumulative frequency column. The cf first equals/exceeds 50 at marks = 30 (cf = 55, having jumped from cf = 15 at marks = 20). So the median lies at marks = 30.
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