NCERT Mathematics · Class 10 · English · 2026-27

Statistics — Exercise 13.1

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

Mean plants per house.

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
0–2 1 1 1
2–4 2 3 6
4–6 1 5 5
6–8 5 7 35
8–10 6 9 54
10–12 2 11 22
12–14 3 13 39

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{162}{20}\approx 8.1\]

Mean ≈ 8.1 plants.

Answer: Approximately 8.1 plants. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q1

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Question 2

Mean daily wage of the 50 workers.

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
500–520 12 510 6120
520–540 14 530 7420
540–560 8 550 4400
560–580 6 570 3420
580–600 10 590 5900

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{27260}{50}\approx 545.2\]

Mean ≈ 545.2 rupees per day.

Answer: Approximately 545.2 rupees per day. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q2

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Question 3

Find the missing pocket-allowance frequency when the mean is ₹18.

Step-by-step solution

Class midpoints are 12,14,16,18,20,22,24. Known frequencies total 7+6+9+13+5+4=44. Their weighted total is 7×12+6×14+9×16+13×18+5×22+4×24=752. The missing class contributes frequency f and product 20f.

\[18=\frac{752+20f}{44+f}\]

\[792+18f=752+20f\;\Rightarrow\;2f=40\]

Answer: f=20.

Resource ID: CB-10-M-CH13-EX13.1-Q3

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Question 4

Mean recorded heartbeats per minute.

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
65–68 2 66.5 133
68–71 4 69.5 278
71–74 3 72.5 217.5
74–77 8 75.5 604
77–80 7 78.5 549.5
80–83 4 81.5 326
83–86 2 84.5 169

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2277}{30}\approx 75.9\]

Mean ≈ 75.9 beats per minute.

Answer: Approximately 75.9 beats per minute. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q4

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Question 5

Mean mango count per box.

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
50–52 15 51 765
53–55 110 54 5940
56–58 135 57 7695
59–61 115 60 6900
62–64 25 63 1575

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{22875}{400}\approx 57.1875\]

Mean ≈ 57.1875 mangoes per box.

Answer: Approximately 57.1875 mangoes per box. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q5

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Question 6

Mean daily food expenditure of 25 households.

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
100–150 4 125 500
150–200 5 175 875
200–250 12 225 2700
250–300 2 275 550
300–350 2 325 650

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{5275}{25}\approx 211\]

Mean ≈ 211 rupees per household.

Answer: Approximately 211 rupees per household. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q6

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Question 7

Mean SO₂ concentration across 30 localities.

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
0–0.04 4 0.02 0.08
0.04–0.08 9 0.06 0.54
0.08–0.12 9 0.1 0.9
0.12–0.16 2 0.14 0.28
0.16–0.2 4 0.18 0.72
0.2–0.24 2 0.22 0.44

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2.96}{30}\approx 0.098667\]

Mean ≈ 0.098667 ppm.

Answer: Approximately 0.098667 ppm. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q7

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Question 8

Mean days absent among 40 students.

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
0–6 11 3 33
6–10 10 8 80
10–14 7 12 84
14–20 4 17 68
20–28 4 24 96
28–38 3 33 99
38–40 1 39 39

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{499}{40}\approx 12.475\]

Mean ≈ 12.475 days.

Answer: Approximately 12.475 days. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q8

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Question 9

Mean literacy rate for 35 cities.

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
45–55 3 50 150
55–65 10 60 600
65–75 11 70 770
75–85 8 80 640
85–95 3 90 270

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2430}{35}\approx 69.428571\]

Mean ≈ 69.428571 percent.

Answer: Approximately 69.428571 percent. Direct midpoint multiplication was used because it is straightforward for this table.

Resource ID: CB-10-M-CH13-EX13.1-Q9

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