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Question
Compute mean from the following data:
| Marks | 0 – 7 | 7 – 14 | 14 – 21 | 21 – 28 | 28 – 35 | 35 – 42 | 42 – 49 |
| Number of Students | 3 | 4 | 7 | 11 | 0 | 16 | 9 |
Solution
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 0 – 7 | 3 | 3 |
| 7 – 14 | 4 | 7 |
| 14 – 21 | 7 | 14 |
| 21 - 28 | 11 | 25 |
| 28 – 35 | 0 | 25 |
| 35 – 42 | 16 | 41 |
| 42 – 49 | 9 | 50 |
| N = Σ𝑓 = 50 |
Now, N = 50 ⇒ `N/2` = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 – 42.
Thus, the median class is 35 – 42.
∴ l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and `N/2` = 25.
Now,
Median =` l +((N/2−c_f)/f) × h`
Median `= 35 + ((25 − 25)/16) xx 7`
= 35 + 0
= 35
Hence, the median age is 35 years.
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