Exercise 13.1 — Statistics

1. 13 marksNCERT Cl-10 Maths, Ex 13.1, Q1

   A survey collected data on the number of plants in 20 houses: 0-2 (1 house), 2-4 (2), 4-6 (1), 6-8 (5), 8-10 (6), 10-12 (2), 12-14 (3). Find the mean number of plants per house, and state which method you used and why.

   Hint. The class marks here (1, 3, 5, ...) are small numbers — that's the signal for which method to reach for.

   Step 1 — Find the class marks (midpoints). 1, 3, 5, 7, 9, 11, 13

   Step 2 — Since these class marks are already small, the direct method is the natural choice — there's no need to shift to an assumed mean just to keep the arithmetic manageable. Σfᵢ = 1+2+1+5+6+2+3 = 20 Σfᵢxᵢ = 1(1)+2(3)+1(5)+5(7)+6(9)+2(11)+3(13) = 1+6+5+35+54+22+39 = 162

   Step 3 — Divide. mean = 162/20 = 8.1

   ✦ Answer: 8.1 plants per house — found by the direct method, since the class marks were small enough that no simplification was needed.

   Where students slip. Reaching for the assumed-mean method out of habit. The method is a choice driven by the size of the numbers involved — with class marks under 15, direct is both correct and quicker.

   Another way. You could still use the assumed-mean method with, say, a = 7: dᵢ = xᵢ − 7 gives −6,−4,−2,0,2,4,6, and Σfᵢdᵢ = 1(−6)+2(−4)+1(−2)+5(0)+6(2)+2(4)+3(6) = −6−8−2+0+12+8+18 = 22, giving mean = 7 + 22/20 = 8.1 — the same answer, more work.

2. 23 marksNCERT Cl-10 Maths, Ex 13.1, Q2

   Find the mean daily wages of 50 workers from: 500-520 (12 workers), 520-540 (14), 540-560 (8), 560-580 (6), 580-600 (10).

   Hint. The class marks are all above 500 — this is exactly when an assumed mean keeps the numbers small.

   Step 1 — Read off the class marks from each wage bracket. 510, 530, 550, 570, 590

   Step 2 — These are large numbers, so take an assumed mean near the centre, a = 550, with class width h = 20. dᵢ = xᵢ − 550: −40, −20, 0, 20, 40

   Step 3 — Compute Σfᵢdᵢ. 12(−40) + 14(−20) + 8(0) + 6(20) + 10(40) = −480 − 280 + 0 + 120 + 400 = −240

   Step 4 — Σfᵢ = 12+14+8+6+10 = 50

   Step 5 — Apply the assumed-mean formula. mean = 550 + (−240/50) = 550 − 4.8

   ✦ Answer: ₹545.20

   Where students slip. Forgetting the sign on a negative Σfᵢdᵢ, turning a subtraction into an addition and reporting 554.8 instead of 545.2.

   Another way. Step-deviation also works here since h = 20 divides every dᵢ evenly: uᵢ = dᵢ/20 gives −2,−1,0,1,2, and Σfᵢuᵢ = 12(−2)+14(−1)+0+6(1)+10(2) = −24−14+6+20 = −12, so mean = 550 + 20(−12/50) = 550 − 4.8 = 545.2 — same answer, smaller numbers to add.

3. 34 marksNCERT Cl-10 Maths, Ex 13.1, Q3

   The daily pocket allowance of children in a locality is given by: 11-13 (7 children), 13-15 (6), 15-17 (9), 17-19 (13), 19-21 (f, unknown), 21-23 (5), 23-25 (4). If the mean allowance is ₹18, find the missing frequency f.

   Hint. Write the mean equation with f left as an unknown, then solve for it — the mean formula becomes a linear equation in f.

   Step 1 — List the class marks. 12, 14, 16, 18, 20, 22, 24

   Step 2 — Compute Σfᵢxᵢ and Σfᵢ in terms of f. Known part: 7(12) + 6(14) + 9(16) + 13(18) + 5(22) + 4(24) = 84 + 84 + 144 + 234 + 110 + 96 = 752 Plus the unknown term: f(20)

   Σfᵢ = 7+6+9+13+f+5+4 = 44 + f

   Step 3 — Set the direct-method mean equal to 18. (752 + 20f) / (44 + f) = 18

   Step 4 — Clear the denominator and solve. 752 + 20f = 18(44 + f) = 792 + 18f 20f − 18f = 792 − 752 2f = 40 f = 20

   Check. With f = 20, Σfᵢ = 64 and Σfᵢxᵢ = 752 + 400 = 1152, so mean = 1152/64 = 18 ✓

   ✦ Answer: f = 20

   Where students slip. Substituting the class mark 20 (for 19-21) as if it were the frequency, confusing which column the unknown belongs to. The unknown f is a *frequency*, not a class mark.

   Another way. Use the assumed-mean method with a = 18 instead: then dᵢ = −6,−4,−2,0,2,4,6, and the mean condition becomes Σfᵢdᵢ = 0 exactly (since a = mean here). That gives 7(−6)+6(−4)+9(−2)+13(0)+f(2)+5(4)+4(6) = 0, i.e. −42−24−18+0+2f+20+24=0, so 2f = 40, f = 20 — the zero-sum trick shortens the algebra.

4. 43 marksNCERT Cl-10 Maths, Ex 13.1, Q4

   Thirty women's heartbeats per minute were recorded: 65-68 (2), 68-71 (4), 71-74 (3), 74-77 (8), 77-80 (7), 80-83 (4), 83-86 (2). Find the mean, choosing a suitable method.

   Hint. The class width here is 3, and the class marks are moderately large — try the assumed-mean or step-deviation method.

   Step 1 — Find the class marks (width 3 throughout). 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5

   Step 2 — Take assumed mean a = 75.5, class width h = 3. uᵢ = (xᵢ − 75.5)/3: −3, −2, −1, 0, 1, 2, 3

   Step 3 — Compute Σfᵢuᵢ. 2(−3) + 4(−2) + 3(−1) + 8(0) + 7(1) + 4(2) + 2(3) = −6 − 8 − 3 + 0 + 7 + 8 + 6 = 4

   Step 4 — Σfᵢ = 2+4+3+8+7+4+2 = 30

   Step 5 — Apply the step-deviation formula, since that is why a small, positive Σfᵢuᵢ nudges the mean just slightly above the assumed value. mean = 75.5 + 3 × (4/30) = 75.5 + 0.4

   ✦ Answer: 75.9 heartbeats per minute

   Where students slip. Using the class boundaries (65, 68, 71, ...) instead of the class marks (66.5, 69.5, ...) when computing uᵢ. The formula always works on the midpoint of each class.

   Another way. Direct method also works, just with more arithmetic: Σfᵢxᵢ = 2(66.5)+4(69.5)+3(72.5)+8(75.5)+7(78.5)+4(81.5)+2(84.5) = 2277, and 2277/30 = 75.9 — same answer, larger numbers throughout.

5. 54 marksNCERT Cl-10 Maths, Ex 13.1, Q5

   The number of mangoes in packing boxes is distributed as: 50-52 (15 boxes), 53-55 (110), 56-58 (135), 59-61 (115), 62-64 (25). Find the mean number of mangoes per box.

   Hint. These classes look continuous but have gaps (52 to 53, for instance) — that doesn't change the class marks, only how you'd handle the median or mode later.

   Step 1 — Find the class marks. Even though the classes are written as 50-52, 53-55, etc. (with small gaps), the class mark is still the midpoint of each printed interval. 51, 54, 57, 60, 63

   Step 2 — Take assumed mean a = 57, class width h = 3. uᵢ = (xᵢ − 57)/3: −2, −1, 0, 1, 2

   Step 3 — Compute Σfᵢuᵢ. 15(−2) + 110(−1) + 135(0) + 115(1) + 25(2) = −30 − 110 + 0 + 115 + 50 = 25

   Step 4 — Σfᵢ = 15+110+135+115+25 = 400

   Step 5 — Apply the step-deviation formula, since Σfᵢuᵢ came out positive, the mean should land a little above the assumed value of 57 — which is exactly what happens. mean = 57 + 3 × (25/400) = 57 + 0.1875

   ✦ Answer: ≈57.19 mangoes per box

   Where students slip. Trying to 'fix' the gaps between classes (52 to 53) by adjusting the class marks with a continuity correction. That adjustment is needed only for the median and mode formulas, which depend on exact class boundaries — the mean, built from class marks, is unaffected.

   Another way. With 400 as such a large Σfᵢ, the direct method would involve four-digit sums throughout — step-deviation's small uᵢ values (−2 to 2) are clearly the better choice here, which is itself worth stating as your reason for choosing the method.

6. 63 marksNCERT Cl-10 Maths, Ex 13.1, Q6

   The daily expenditure on food of 25 households is: 100-150 (4), 150-200 (5), 200-250 (12), 250-300 (2), 300-350 (2). Find the mean daily expenditure by a suitable method.

   Hint. Class width is 50 here — a clean common factor for every deviation.

   Step 1 — Determine the class marks. 125, 175, 225, 275, 325

   Step 2 — Take assumed mean a = 225, class width h = 50. uᵢ = (xᵢ − 225)/50: −2, −1, 0, 1, 2

   Step 3 — Compute Σfᵢuᵢ. 4(−2) + 5(−1) + 12(0) + 2(1) + 2(2) = −8 − 5 + 0 + 2 + 4 = −7

   Step 4 — Σfᵢ = 4+5+12+2+2 = 25

   Step 5 — Apply the step-deviation formula. Since Σfᵢuᵢ is negative, the mean must come out below the assumed value of 225, which is why the last step is a subtraction. mean = 225 + 50 × (−7/25) = 225 − 14

   ✦ Answer: ₹211

   Where students slip. Losing track of the negative sign in −7/25 and adding 14 instead of subtracting it, landing on 239 rather than 211.

   Another way. Check the plausibility: with the highest frequency (12) sitting in the 200-250 class, a mean of 211 landing inside that very class is exactly what you'd expect — a quick sanity check worth doing on every mean-of-grouped-data answer.

7. 74 marksNCERT Cl-10 Maths, Ex 13.1, Q7

   The SO₂ concentration (in ppm) for 30 localities is: 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). Find the mean concentration.

   Hint. The class width here is a small decimal (0.04) — the method still works exactly the same way, just with smaller numbers.

   Step 1 — Set out the class marks. 0.02, 0.06, 0.10, 0.14, 0.18, 0.22

   Step 2 — Take assumed mean a = 0.14, class width h = 0.04. uᵢ = (xᵢ − 0.14)/0.04: −3, −2, −1, 0, 1, 2

   Step 3 — Compute Σfᵢuᵢ. 4(−3) + 9(−2) + 9(−1) + 2(0) + 4(1) + 2(2) = −12 − 18 − 9 + 0 + 4 + 4 = −31

   Step 4 — Σfᵢ = 4+9+9+2+4+2 = 30

   Step 5 — Apply the step-deviation formula. Because Σfᵢuᵢ is negative here, the mean ends up a little below the assumed value of 0.14. mean = 0.14 + 0.04 × (−31/30) = 0.14 − 0.04133...

   ✦ Answer: ≈0.0987 ppm (about 0.099 ppm)

   Where students slip. Treating the decimal class width as if it needed different handling from a whole-number one. The step-deviation method doesn't care whether h is 20 or 0.04 — the formula is identical.

   Another way. Direct method, for comparison: Σfᵢxᵢ = 4(0.02)+9(0.06)+9(0.10)+2(0.14)+4(0.18)+2(0.22) = 0.08+0.54+0.90+0.28+0.72+0.44 = 2.96, and 2.96/30 ≈ 0.0987 — same answer, confirming the step-deviation result.

8. 84 marksNCERT Cl-10 Maths, Ex 13.1, Q8

   A class teacher's absentee record for 40 students over a term is: 0-6 (11), 6-10 (10), 10-14 (7), 14-20 (4), 20-28 (4), 28-38 (3), 38-40 (1). Find the mean number of days absent.

   Hint. The class widths are different from one class to the next — step-deviation with a single h no longer applies cleanly, so fall back on assumed mean.

   Step 1 — Find the class marks. Even with unequal widths, each mark is still the midpoint of its own class. 0-6 → 3, 6-10 → 8, 10-14 → 12, 14-20 → 17, 20-28 → 24, 28-38 → 33, 38-40 → 39

   Step 2 — Since the class widths vary (6, 4, 4, 6, 8, 10, 2), there is no single h to divide every deviation by cleanly, so use the assumed-mean method instead of step-deviation. Take a = 17. dᵢ = xᵢ − 17: −14, −9, −5, 0, 7, 16, 22

   Step 3 — Compute Σfᵢdᵢ. 11(−14) + 10(−9) + 7(−5) + 4(0) + 4(7) + 3(16) + 1(22) = −154 − 90 − 35 + 0 + 28 + 48 + 22 = −181

   Step 4 — Σfᵢ = 11+10+7+4+4+3+1 = 40

   Step 5 — Apply the assumed-mean formula. mean = 17 + (−181/40) = 17 − 4.525

   ✦ Answer: ≈12.475 days

   Where students slip. Trying to force step-deviation onto unequal class widths by picking one h and dividing every dᵢ by it regardless. Step-deviation's simplification only works when a single h is the actual class width everywhere — here it isn't, so assumed mean is the right tool.

   Another way. Direct method as a check: Σfᵢxᵢ = 11(3)+10(8)+7(12)+4(17)+4(24)+3(33)+1(39) = 33+80+84+68+96+99+39 = 499, and 499/40 = 12.475 ✓ — matches exactly.

9. 93 marksNCERT Cl-10 Maths, Ex 13.1, Q9

   The literacy rate (%) of 35 cities is: 45-55 (3), 55-65 (10), 65-75 (11), 75-85 (8), 85-95 (3). Find the mean literacy rate.

   Hint. Class width 10 throughout — a clean setup for step-deviation.

   Step 1 — Note the class marks. 50, 60, 70, 80, 90

   Step 2 — Take assumed mean a = 70, class width h = 10. uᵢ = (xᵢ − 70)/10: −2, −1, 0, 1, 2

   Step 3 — Compute Σfᵢuᵢ. 3(−2) + 10(−1) + 11(0) + 8(1) + 3(2) = −6 − 10 + 0 + 8 + 6 = −2

   Step 4 — Σfᵢ = 3+10+11+8+3 = 35

   Step 5 — Apply the step-deviation formula. Since Σfᵢuᵢ is negative, the mean lands just under the assumed value of 70, which is why the answer is close to but below 70%. mean = 70 + 10 × (−2/35) = 70 − 4/7

   ✦ Answer: ≈69.43%

   Where students slip. Rounding −2/35 to a clean decimal too early (e.g. −0.06) instead of carrying the fraction through, which introduces a small but avoidable error in the final answer.

   Another way. The distribution is nearly symmetric around 70 (frequencies 3,10,11,8,3), so a mean of about 69.4 — just slightly below 70 — matches the eye-test that the left side has marginally more weight than the right.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter still has three exercises — 13.1 mean with 9 questions, 13.2 mode with 6, 13.3 median with 7; the step-deviation method and cumulative frequency are both retained; only a dedicated ogive-construction exercise is no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.