CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Solutions Class 10 Maths Chapter 13 Statistics
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Statistics is in Statistics and Probability, which carries 11 of the 80 marks in the 2026-27 CBSE Maths paper
7 NCERT textbook questions from Chapter 13
Exercise 13.1
NCERT Ex 13.1 Q23 marks
Find the mean daily wage from: Daily wages (₹) 500-520 has 12 workers, 520-540 has 14, 540-560 has 8, 560-580 has 6, 580-600 has 10. (Use step-deviation, a = 550, h = 20.)
The mean of the following distribution is ₹18. Find the missing frequency f. Daily pocket allowance (₹) 11-13 → 7, 13-15 → 6, 15-17 → 9, 17-19 → 13, 19-21 → f, 21-23 → 5, 23-25 → 4.
For monthly electricity consumption (in units): 65-85 → 4, 85-105 → 5, 105-125 → 13, 125-145 → 20, 145-165 → 14, 165-185 → 8, 185-205 → 4. Find the median, mean and mode of the data and compare them.
Step 1[1 mark] Setup: n = 68, n/2 = 34. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68, so the median class is 125-145 (first cf ≥ 34); l = 125, cf = 22, f = 20, h = 20.
Step-by-step solution
[1 mark] Setup: n = 68, n/2 = 34. Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68, so the median class is 125-145 (first cf ≥ 34); l = 125, cf = 22, f = 20, h = 20.
[1 mark] Median = l + ((n/2 − cf)/f) × h = 125 + ((34 − 22)/20) × 20 = 125 + 12 = 137 units.
[1 mark] Mean by direct method: class marks 75, 95, 115, 135, 155, 175, 195; Σfx = 300 + 475 + 1495 + 2700 + 2170 + 1400 + 780 = 9320.
[1 mark] Mean = Σfx/Σf = 9320/68 ≈ 137.06 units.
[1 mark] Mode: modal class 125-145 (f₁ = 20, f₀ = 13, f₂ = 14, h = 20); Mode = 125 + ((20 − 13)/(40 − 13 − 14)) × 20 = 125 + (7/13) × 20 ≈ 135.77 units. Comparing: Mean (137.06) ≈ Median (137) > Mode (135.77), so the distribution is nearly symmetric.
Answer:Mean ≈ 137.06, Median = 137, Mode ≈ 135.77 — distribution near-symmetric.
The median of the following data is 525 and the total frequency is 100. Find x and y. 0-100 → 2, 100-200 → 5, 200-300 → x, 300-400 → 12, 400-500 → 17, 500-600 → 20, 600-700 → y, 700-800 → 9, 800-900 → 7, 900-1000 → 4.
Step 1[1 mark] The total frequency is 100 and the known frequencies sum to 76, so 76 + x + y = 100 ⇒ x + y = 24 …(i).
Step-by-step solution
[1 mark] The total frequency is 100 and the known frequencies sum to 76, so 76 + x + y = 100 ⇒ x + y = 24 …(i).
[1 mark] Median = 525 lies in the class 500-600, so l = 500, f = 20, h = 100 and n/2 = 50.
[1 mark] Cumulative frequency preceding the median class: cf = 2 + 5 + x + 12 + 17 = 36 + x.
[1 mark] Apply the median formula: 525 = 500 + [(50 − (36 + x))/20] × 100 ⇒ 25 = (14 − x) × 5 ⇒ 14 − x = 5 ⇒ x = 9.
[1 mark] From (i): y = 24 − 9 = 15. Hence x = 9 and y = 15.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Case-based4 marks
A school's environment club surveys SO₂ concentration (in ppm) for 30 localities. Concentration of SO₂: 0.00-0.04 → 4, 0.04-0.08 → 9, 0.08-0.12 → 9, 0.12-0.16 → 2, 0.16-0.20 → 4, 0.20-0.24 → 2.
(i) Find the class mark of the class 0.04-0.08.
(ii) Find Σfᵢ and Σfᵢxᵢ.
(iii) Find the mean SO₂ concentration.
(iv) Which method did you use and why?
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Questions from the NCERT Class 10 Mathematics textbook (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.