NCERT Mathematics · Class 10 · English · 2026-27
Statistics — Exercise 13.2
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
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Question 1
Patient age distribution: find the mode and mean.
Step-by-step solution
Use the direct method: represent each class by its midpoint xᵢ, multiply by fᵢ, and divide the product total by the frequency total. These are grouped-data estimates.
| Class | fᵢ | Midpoint xᵢ | fᵢxᵢ |
|---|---|---|---|
| 5–15 | 6 | 10 | 60 |
| 15–25 | 11 | 20 | 220 |
| 25–35 | 21 | 30 | 630 |
| 35–45 | 23 | 40 | 920 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 5 | 60 | 300 |
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2830}{80}\approx 35.375\]
Mean ≈ 35.375 years.
The modal class is 35–45, with largest frequency 23. Use l=35, h=10, f₁=23, f₀=21, f₂=14.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=35+\frac{2}{11}\times 10\approx 36.818182\]
Modal estimate ≈ 36.818182 years.
The modal age is around 36.82 years in the most frequent class 35–45; the mean around 35.38 averages all ages.
Resource ID: CB-10-M-CH13-EX13.2-Q1
Question 2
Electrical component lifetimes: find the mode.
Step-by-step solution
| Class | Frequency |
|---|---|
| 0–20 | 10 |
| 20–40 | 35 |
| 40–60 | 52 |
| 60–80 | 61 |
| 80–100 | 38 |
| 100–120 | 29 |
The modal class is 60–80, with largest frequency 61. Use l=60, h=20, f₁=61, f₀=52, f₂=38.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=60+\frac{9}{32}\times 20\approx 65.625\]
Modal estimate ≈ 65.625 hours.
Resource ID: CB-10-M-CH13-EX13.2-Q2
Question 3
Monthly expenditure of 200 families: find the mode and mean.
Step-by-step solution
Use the direct method: represent each class by its midpoint xᵢ, multiply by fᵢ, and divide the product total by the frequency total. These are grouped-data estimates.
| Class | fᵢ | Midpoint xᵢ | fᵢxᵢ |
|---|---|---|---|
| 1000–1500 | 24 | 1250 | 30000 |
| 1500–2000 | 40 | 1750 | 70000 |
| 2000–2500 | 33 | 2250 | 74250 |
| 2500–3000 | 28 | 2750 | 77000 |
| 3000–3500 | 30 | 3250 | 97500 |
| 3500–4000 | 22 | 3750 | 82500 |
| 4000–4500 | 16 | 4250 | 68000 |
| 4500–5000 | 7 | 4750 | 33250 |
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{532500}{200}\approx 2662.5\]
Mean ≈ 2662.5 rupees per month.
The modal class is 1500–2000, with largest frequency 40. Use l=1500, h=500, f₁=40, f₀=24, f₂=33.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=1500+\frac{16}{23}\times 500\approx 1847.826087\]
Modal estimate ≈ 1847.826087 rupees per month.
The most frequent spending class is ₹1500–2000; the mean also reflects families spending larger amounts.
Resource ID: CB-10-M-CH13-EX13.2-Q3
Question 4
The textbook’s historical teacher-student ratio distribution: find the mode and mean.
Step-by-step solution
Use the direct method: represent each class by its midpoint xᵢ, multiply by fᵢ, and divide the product total by the frequency total. These are grouped-data estimates.
| Class | fᵢ | Midpoint xᵢ | fᵢxᵢ |
|---|---|---|---|
| 15–20 | 3 | 17.5 | 52.5 |
| 20–25 | 8 | 22.5 | 180 |
| 25–30 | 9 | 27.5 | 247.5 |
| 30–35 | 10 | 32.5 | 325 |
| 35–40 | 3 | 37.5 | 112.5 |
| 40–45 | 0 | 42.5 | 0 |
| 45–50 | 0 | 47.5 | 0 |
| 50–55 | 2 | 52.5 | 105 |
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1022.5}{35}\approx 29.214286\]
Mean ≈ 29.214286 students per teacher.
The modal class is 30–35, with largest frequency 10. Use l=30, h=5, f₁=10, f₀=9, f₂=3.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=30+\frac{1}{8}\times 5\approx 30.625\]
Modal estimate ≈ 30.625 students per teacher.
The modal estimate describes the most frequent class of states/UTs, while the mean averages the grouped state ratios; this is supplied exercise data, not a current national statistic.
Resource ID: CB-10-M-CH13-EX13.2-Q4
Question 5
Runs scored by the listed batsmen: find the mode.
Step-by-step solution
| Class | Frequency |
|---|---|
| 3000–4000 | 4 |
| 4000–5000 | 18 |
| 5000–6000 | 9 |
| 6000–7000 | 7 |
| 7000–8000 | 6 |
| 8000–9000 | 3 |
| 9000–10000 | 1 |
| 10000–11000 | 1 |
The modal class is 4000–5000, with largest frequency 18. Use l=4000, h=1000, f₁=18, f₀=4, f₂=9.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=4000+\frac{14}{23}\times 1000\approx 4608.695652\]
Modal estimate ≈ 4608.695652 runs.
Resource ID: CB-10-M-CH13-EX13.2-Q5
Question 6
Car counts in 100 three-minute observation periods: find the mode.
Step-by-step solution
| Class | Frequency |
|---|---|
| 0–10 | 7 |
| 10–20 | 14 |
| 20–30 | 13 |
| 30–40 | 12 |
| 40–50 | 20 |
| 50–60 | 11 |
| 60–70 | 15 |
| 70–80 | 8 |
The modal class is 40–50, with largest frequency 20. Use l=40, h=10, f₁=20, f₀=12, f₂=11.
\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=40+\frac{8}{17}\times 10\approx 44.705882\]
Modal estimate ≈ 44.705882 cars per three minutes.
Resource ID: CB-10-M-CH13-EX13.2-Q6