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Question
Find the mean:
| Marks (below) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| No. of students | 5 | 9 | 17 | 29 | 45 | 60 | 70 | 78 | 83 | 85 |
Sum
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Solution
Step 1: Create a frequency distribution table
The “Marks (below)” represent the upper limits of class intervals. Since they increase by 10, our intervals will be 0 – 10, 10 – 20, etc. To find the frequency (f) for each interval, we subtract the previous cumulative frequency from the current one.
| Class Interval (Marks) |
Midpoint (xi) |
No. of students (cf) |
Frequency (fi) |
fi × xi |
| 0 – 10 | 5 | 5 | 5 | 25 |
| 10 – 20 | 15 | 9 | 9 – 5 = 4 | 60 |
| 20 – 30 | 25 | 17 | 17 – 9 = 8 | 200 |
| 30 – 40 | 35 | 29 | 29 – 17 = 12 | 420 |
| 40 – 50 | 45 | 45 | 45 – 29 = 16 | 720 |
| 50 – 60 | 55 | 60 | 60 – 45 = 15 | 825 |
| 60 – 70 | 65 | 70 | 70 – 60 = 10 | 650 |
| 70 – 80 | 75 | 78 | 78 – 70 = 8 | 600 |
| 80 – 90 | 85 | 83 | 83 – 78 = 5 | 425 |
| 90 – 100 | 95 | 85 | 85 – 83 = 2 | 190 |
| Total | Σfi = 85 | Σfixi = 4115 |
Step 2: Calculate the mean
The formula for the mean `(barx)` of grouped data is:
`barx = (sumf_ix_i)/(sumf_i)`
Substituting the values from our table:
`barx = 4115/85`
`barx ≈ 48.41`
The mean marks of the students is approximately 48.41.
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