Question of 20
Q.
Compute mode of the following series :
| Mid values | Frequency |
|---|---|
| 5 | 7 |
| 10 | 13 |
| 15 | 19 |
| 20 | 24 |
| 25 | 32 |
| 30 | 28 |
| 35 | 17 |
| 40 | 8 |
| 45 | 6 |
OR
Calculate arithmetic mean of the following series by using Step Deviation Method :
| Class Interval | Frequency |
|---|---|
| 5-15 | 8 |
| 15-25 | 12 |
| 25-35 | 15 |
| 35-45 | 9 |
| 45-55 | 6 |
Jammu Kashmir JkboseJKBOSE Class 11 (Commerce) 2025Subjective· 4mImportance★★★★★est
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Start your 14-day free trial to unlock the full solution →Mode of the mid-value series ≈ 25.83.
Step 1 — Convert mid-values to class intervals. Since consecutive mid-values differ by 5, the class width , and each class boundary is (mid-value ± 2.5):
| Mid value | Class Interval | Frequency (f) |
|---|---|---|
| 5 | 2.5–7.5 | 7 |
| 10 | 7.5–12.5 | 13 |
| 15 | 12.5–17.5 | 19 |
| 20 | 17.5–22.5 | 24 |
| 25 | 22.5–27.5 | 32 ← highest frequency |
| 30 | 27.5–32.5 | 28 |
| 35 | 32.5–37.5 | 17 |
| 40 | 37.5–42.5 | 8 |
| 45 | 42.5–47.5 | 6 |
Step 2 — Identify the modal class. The highest frequency is 32, at the class 22.5–27.5. So:
(lower limit of modal class), (modal class frequency), (frequency of class preceding modal class), (frequency of class succeeding modal class), .
Step 3 — Apply the mode formula:
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