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.

Answer: Mode ≈ 36.818182 years; mean ≈ 35.375 years.

Resource ID: CB-10-M-CH13-EX13.2-Q1

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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.

Answer: Mode ≈ 65.625 hours.

Resource ID: CB-10-M-CH13-EX13.2-Q2

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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.

Answer: Mode ≈ 1847.826087 rupees per month; mean ≈ 2662.5 rupees per month.

Resource ID: CB-10-M-CH13-EX13.2-Q3

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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.

Answer: Mode ≈ 30.625 students per teacher; mean ≈ 29.214286 students per teacher.

Resource ID: CB-10-M-CH13-EX13.2-Q4

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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.

Answer: Mode ≈ 4608.695652 runs.

Resource ID: CB-10-M-CH13-EX13.2-Q5

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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.

Answer: Mode ≈ 44.705882 cars per three minutes.

Resource ID: CB-10-M-CH13-EX13.2-Q6

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