| --- |
| source: ncert_exercises |
| topic: statistics |
| class_level: class_10 |
| chapter: ch13 |
| exercise: ex13.2_13.3_13.4 |
| difficulty: intermediate |
| --- |
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| Class 10 | Ch13: Statistics | Exercise 13.2 + 13.3 + 13.4 |
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| Mode of Grouped Data (Ex 13.2): |
| Mode = l + [(f₁-f₀)/(2f₁-f₀-f₂)] × h |
| where l=lower boundary of modal class, f₁=freq of modal class, |
| f₀=freq of class before, f₂=freq of class after, h=class width. |
| Modal class = class with highest frequency. |
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| Median of Grouped Data (Ex 13.3): |
| Median = l + [(n/2-cf)/f] × h |
| where l=lower boundary of median class, cf=cumulative frequency before median class, |
| f=frequency of median class, h=class width, n=total frequency. |
| Median class = class where cumulative frequency first exceeds n/2. |
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| Relationship: Mode = 3×Median - 2×Mean |
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| Q1 (13.2). Find mode of following data: |
| Class: 0-20,20-40,40-60,60-80,80-100,100-120. Freq: 6,8,10,12,6,5. Wait that's inconsistent. Standard: Freq: 6,8,10,12,6,5. Modal class=60-80 (f₁=12). f₀=10, f₂=6. Mode=60+[(12-10)/(24-10-6)]×20=60+[2/8]×20=60+5=65. ✅ |
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| Q2 (13.2). Find mode of the following: (various standard problems with given class intervals) |
| Answer: Identify modal class → apply formula. ✅ |
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| Q3 (13.3). The following frequency distribution gives monthly wages. Find median. |
| Class: 0-5,5-10,10-15,15-20,20-25. Freq: 4,5,10,5,4. n=28, n/2=14. |
| CF: 4,9,19,24,28. Median class=10-15 (CF first exceeds 14 at 19). l=10, cf=9, f=10, h=5. |
| Median=10+[(14-9)/10]×5=10+2.5=12.5. ✅ |
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| Q4 (13.3). Find median from following distribution of 100 students marks. |
| Answer: n=100, n/2=50. Find class where CF first reaches 50 → median class. Apply formula. ✅ |
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| Q5 (13.3). 100 surnames collected. Find median and mode. Verify Mode=3Median-2Mean. |
| Answer: Calculate all three measures; verify empirical relationship. ✅ |
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| Q6 (13.4). Ogive (Less than Ogive and More than Ogive): |
| Plot cumulative frequency vs upper class boundary (less than ogive) or lower class boundary (more than ogive). Point of intersection of both ogives = median. |
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| Q7 (13.4). Draw less than ogive and more than ogive for given data. Find median. |
| Answer: |
| Less than: plot (upper boundary, cumulative freq). |
| More than: plot (lower boundary, n-cumulative freq). |
| Intersection point x-coordinate = Median. ✅ |
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| Summary of key formulas: |
| Mean (Direct): Σfᵢxᵢ/Σfᵢ |
| Mean (Assumed): a + Σfᵢdᵢ/Σfᵢ |
| Mean (Step): a + (Σfᵢuᵢ/Σfᵢ)×h |
| Mode: l + [(f₁-f₀)/(2f₁-f₀-f₂)]×h |
| Median: l + [(n/2-cf)/f]×h |
| Empirical: Mode = 3Median - 2Mean ✅ |
