Exercise 13.2 — Statistics

1. 14 marksNCERT Cl-10 Maths, Ex 13.2, Q1

   Ages of patients admitted in a hospital during a year: 5-15 (6), 15-25 (11), 25-35 (21), 35-45 (23), 45-55 (14), 55-65 (5). Find the mode and the mean of the data, and compare and interpret the two measures.

   Hint. The modal class is simply wherever the frequency is highest — no other reasoning needed to locate it.

   Finding the mode.

   Step 1 — Find the modal class: the one with the highest frequency. Frequencies are 6, 11, 21, 23, 14, 5 — the highest is 23, in class 35-45.

   Step 2 — Read off the values needed. l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10

   Step 3 — Apply the mode formula, since f₁ is only slightly larger than f₀ here, the correction term should be small — which is exactly why the mode lands close to l. Mode = 35 + [(23−21)/(2×23−21−14)] × 10 = 35 + [2/11] × 10

   Step 4 — Compute. = 35 + 20/11 ≈ 35 + 1.818

   ✦ Mode ≈ 36.8 years

   Mean.

   Step 5 — Class marks: 10, 20, 30, 40, 50, 60. Using the assumed mean a = 40, h = 10: uᵢ = (xᵢ−40)/10: −3, −2, −1, 0, 1, 2 Σfᵢuᵢ = 6(−3)+11(−2)+21(−1)+23(0)+14(1)+5(2) = −18−22−21+0+14+10 = −37 Σfᵢ = 80 mean = 40 + 10(−37/80) = 40 − 4.625

   ✦ Mean ≈ 35.4 years

   Interpretation. The maximum number of patients admitted are around 36.8 years old (the mode), while the average age of all patients is about 35.4 years — the two are close, telling you the data doesn't have a strongly skewed tail pulling the average away from the most common age.

   Where students slip. Choosing the modal class by eye rather than strictly by 'highest frequency' — with several classes having double-digit frequencies (11, 21, 23, 14), it's easy to grab a plausible-looking one instead of checking all six numbers.

   Another way. The empirical relationship offers a rough cross-check: 3×median − 2×mean should be near the mode. Computing the median here (not asked, but a private check) would land close to 36, consistent with the mode of 36.8.

2. 23 marksNCERT Cl-10 Maths, Ex 13.2, Q2

   The observed lifetimes (in hours) of 225 electrical components are: 0-20 (10), 20-40 (35), 40-60 (52), 60-80 (61), 80-100 (38), 100-120 (29). Determine the modal lifetime of the components.

   Hint. Same procedure as Q1 — just find the modal class and substitute.

   Step 1 — Identify the modal class. Frequencies: 10, 35, 52, 61, 38, 29 — the highest is 61, in class 60-80.

   Step 2 — Read off the values. l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20

   Step 3 — Apply the formula. Since f₁ exceeds both neighbours by a healthy margin, the mode should sit meaningfully above l, which is why the correction term isn't negligible here. Mode = 60 + [(61−52)/(2×61−52−38)] × 20 = 60 + [9/32] × 20

   Step 4 — Compute. = 60 + 180/32 = 60 + 5.625

   ✦ Answer: 65.625 hours

   Where students slip. Using f₀ and f₂ from the wrong neighbouring classes — f₀ must be the class *immediately before* the modal class (40-60, frequency 52) and f₂ the one *immediately after* (80-100, frequency 38), not any other pair.

   Another way. Sanity-check the direction of the shift: since f₁ > f₀ (61 > 52) but f₁ is also well above f₂ (61 vs 38), the mode should land noticeably above l = 60 but well short of l + h = 80 — 65.6 fits that expectation.

3. 34 marksNCERT Cl-10 Maths, Ex 13.2, Q3

   The distribution of monthly household expenditure for 200 families: 1000-1500 (24), 1500-2000 (40), 2000-2500 (33), 2500-3000 (28), 3000-3500 (30), 3500-4000 (22), 4000-4500 (16), 4500-5000 (7). Find the modal monthly expenditure, and also the mean monthly expenditure.

   Hint. Eight classes this time, but the same two procedures — mode from the highest frequency, mean by assumed mean or step-deviation.

   Finding the mode first.

   Step 1 — Find the modal class. Frequencies: 24, 40, 33, 28, 30, 22, 16, 7 — highest is 40, in class 1500-2000.

   Step 2 — Read off the values. l = 1500, f₁ = 40, f₀ = 24, f₂ = 33, h = 500

   Step 3 — Apply the formula, since the numerator (16) is a large fraction of the denominator (23), the mode should sit well into the modal class rather than near its start. Mode = 1500 + [(40−24)/(2×40−24−33)] × 500 = 1500 + [16/23] × 500

   Step 4 — Compute. = 1500 + 8000/23 ≈ 1500 + 347.8

   ✦ Mode ≈ ₹1847.83

   Mean.

   Step 5 — Class marks: 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750. Assumed mean a = 2750, h = 500: uᵢ: −3, −2, −1, 0, 1, 2, 3, 4 Σfᵢuᵢ = 24(−3)+40(−2)+33(−1)+28(0)+30(1)+22(2)+16(3)+7(4) = −72−80−33+0+30+44+48+28 = −35 Σfᵢ = 24+40+33+28+30+22+16+7 = 200 mean = 2750 + 500(−35/200) = 2750 − 87.5

   ✦ Mean ≈ ₹2662.50

   Where students slip. Miscounting how many classes there are and misaligning uᵢ values, especially with eight classes to track instead of the more common five or six — write out the uᵢ column against the class list explicitly before summing.

   Another way. Compute Σfᵢ as a running check while building the uᵢ column: 24, 64, 97, 125, 155, 177, 193, 200 — the final running total should exactly match the stated 200 families, catching any dropped or duplicated row.

4. 44 marksNCERT Cl-10 Maths, Ex 13.2, Q4

   The state-wise teacher-student ratio in higher secondary schools of India: 15-20 (3), 20-25 (8), 25-30 (9), 30-35 (10), 35-40 (3), 40-45 (0), 45-50 (0), 50-55 (2). Find the mode and mean, and interpret the two measures.

   Hint. Two of the classes have zero frequency — treat them like any other class when reading off f₀ and f₂, they simply contribute 0.

   Mode, computed first.

   Step 1 — Find the modal class. Frequencies: 3, 8, 9, 10, 3, 0, 0, 2 — highest is 10, in class 30-35.

   Step 2 — Read off the values. l = 30, f₁ = 10, f₀ = 9, f₂ = 3, h = 5

   Step 3 — Apply the formula. Because f₁ and f₀ are nearly equal (10 versus 9), the correction stays tiny, which is why the mode barely moves past l. Mode = 30 + [(10−9)/(2×10−9−3)] × 5 = 30 + [1/8] × 5

   Step 4 — Compute. = 30 + 5/8 = 30.625

   ✦ Mode ≈ 30.6

   Mean.

   Step 5 — Class marks: 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5. Assumed mean a = 32.5, h = 5: uᵢ: −3, −2, −1, 0, 1, 2, 3, 4 Σfᵢuᵢ = 3(−3)+8(−2)+9(−1)+10(0)+3(1)+0(2)+0(3)+2(4) = −9−16−9+0+3+0+0+8 = −23 Σfᵢ = 3+8+9+10+3+0+0+2 = 35 mean = 32.5 + 5(−23/35) ≈ 32.5 − 3.286

   ✦ Mean ≈ 29.2

   Interpretation. The most common teacher-student ratio across states is about 30.6, while the overall average across all states is a little lower, about 29.2 — most states cluster near 30-35, with the mean pulled down slightly by the smaller number of states with lower ratios.

   Where students slip. Skipping the two zero-frequency classes entirely, as if they didn't exist, which shifts every subsequent uᵢ value by one position and throws off the whole sum.

   Another way. Keep the zero-frequency classes in the table but mentally note their contribution is always 0 regardless of uᵢ — that keeps the row count consistent without needing to actually compute 0 × anything by hand each time.

5. 53 marksNCERT Cl-10 Maths, Ex 13.2, Q5

   The distribution of runs scored by top ODI batsmen: 3000-4000 (4), 4000-5000 (18), 5000-6000 (9), 6000-7000 (7), 7000-8000 (6), 8000-9000 (3), 9000-10000 (1), 10000-11000 (1). Find the mode of the data.

   Hint. The highest frequency here is much larger than the rest — the modal class should be obvious at a glance.

   Step 1 — Scan for the modal class. Frequencies: 4, 18, 9, 7, 6, 3, 1, 1 — the highest is 18, clearly in class 4000-5000, well above every other class.

   Step 2 — Read off the values. l = 4000, f₁ = 18, f₀ = 4, f₂ = 9, h = 1000

   Step 3 — Apply the formula. Since f₁ dominates both neighbours so heavily, the correction term is large, which is why the mode sits well past the halfway point of its class. Mode = 4000 + [(18−4)/(2×18−4−9)] × 1000 = 4000 + [14/23] × 1000

   Step 4 — Compute. = 4000 + 14000/23 ≈ 4000 + 608.7

   ✦ Answer: ≈4608.70 runs

   Where students slip. Forgetting that f₀ is the class *before* 4000-5000, which is 3000-4000 (frequency 4) — not zero, since 3000-4000 is the very first class listed and does have a frequency.

   Another way. Because f₁ (18) so heavily outweighs both neighbours (4 and 9), the numerator (14) is a large fraction of the denominator (23), so the mode should land well over halfway through the modal class — 608.7 out of a possible 1000 confirms this.

6. 63 marksNCERT Cl-10 Maths, Ex 13.2, Q6

   The number of cars passing a spot in 100 three-minute periods: 0-10 (7), 10-20 (14), 20-30 (13), 30-40 (12), 40-50 (20), 50-60 (11), 60-70 (15), 70-80 (8). Find the mode of the data.

   Hint. With eight classes and frequencies that don't jump out immediately, scan the whole list carefully before deciding the modal class.

   Step 1 — Locate the modal class. Frequencies: 7, 14, 13, 12, 20, 11, 15, 8 — the highest is 20, in class 40-50. (It's easy to be distracted by 14, 13 or 15 nearby, but 20 is the true maximum.)

   Step 2 — Read off the values. l = 40, f₁ = 20, f₀ = 12, f₂ = 11, h = 10

   Step 3 — Apply the formula, since f₁ clears f₀ by a wide margin, the mode should land solidly past the halfway mark of its class. Mode = 40 + [(20−12)/(2×20−12−11)] × 10 = 40 + [8/17] × 10

   Step 4 — Compute. = 40 + 80/17 ≈ 40 + 4.706

   ✦ Answer: ≈44.71 cars

   Where students slip. Picking class 60-70 (frequency 15) as the modal class because it's the second-highest and close to other high values, without checking that 40-50's frequency of 20 is actually larger.

   Another way. List the frequencies in a column next to their classes and circle the maximum before doing anything else — a one-second visual check that prevents exactly the error above.

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.