Exercise 13.3 — Statistics

1. 15 marksNCERT Cl-10 Maths, Ex 13.3, Q1

   The monthly electricity consumption of 68 consumers: 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.

   Hint. All three measures are wanted this time — build the cumulative frequency column once, and it serves the median directly while mode and mean use the frequencies alone.

   Finding the median.

   Step 1 — Build the cumulative frequency column. 4, 9, 22, 42, 56, 64, 68

   Step 2 — Find n/2. n = 68, so n/2 = 34

   Step 3 — Find the first class whose cumulative frequency reaches 34. 22 (up to 105-125) is short of 34; 42 (up to 125-145) reaches it. So the median class is 125-145, with cf = 22 (the cumulative frequency of the class before it).

   Step 4 — Read off the values and apply the formula. l = 125, cf = 22, f = 20, h = 20 Median = 125 + [(34−22)/20] × 20 = 125 + 12

   ✦ Median = 137

   Mode.

   Step 5 — The modal class is the one with the highest frequency: 20, in class 125-145 — the same class as the median, here. l = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20 Mode = 125 + [(20−13)/(2×20−13−14)] × 20 = 125 + [7/13] × 20 ≈ 125 + 10.77

   ✦ Mode ≈ 135.77

   Mean.

   Step 6 — Class marks: 75, 95, 115, 135, 155, 175, 195. Assumed mean a = 135, h = 20: uᵢ: −3, −2, −1, 0, 1, 2, 3 Σfᵢuᵢ = 4(−3)+5(−2)+13(−1)+20(0)+14(1)+8(2)+4(3) = −12−10−13+0+14+16+12 = 7 mean = 135 + 20(7/68) ≈ 135 + 2.06

   ✦ Mean ≈ 137.06

   Comparison. All three measures land within about 1.3 units of each other (135.77, 137, 137.06) — a sign of a fairly symmetric distribution, with no long tail dragging any one measure far from the others.

   Where students slip. Computing all three in isolation without noticing they land close together, missing the chance to catch an arithmetic slip — if one of the three had come out wildly different, that would be the signal to recheck it.

   Another way. Check with the empirical relationship: Mode ≈ 3×Median − 2×Mean = 3(137) − 2(137.06) = 411 − 274.12 = 136.88, close to the computed 135.77 — near enough to confirm no major error, given the relationship is approximate.

2. 24 marksNCERT Cl-10 Maths, Ex 13.3, Q2

   If the median of the following distribution is 28.5, find x and y: 0-10 (5), 10-20 (x), 20-30 (20), 30-40 (15), 40-50 (y), 50-60 (5); total frequency 60.

   Hint. The median value (28.5) tells you exactly which class the median falls in — use that to write one equation, and the total frequency to write a second.

   Step 1 — Use the total to write one equation. 5 + x + 20 + 15 + y + 5 = 60 x + y = 15 … (1)

   Step 2 — Identify the median class from the given median value. 28.5 falls within 20-30, so that is the median class: l = 20, f = 20, h = 10. The cumulative frequency before this class is 5 + x.

   Step 3 — Write n/2. Since the total is 60, n/2 = 30.

   Step 4 — Apply the median formula and set it equal to 28.5. 28.5 = 20 + [(30 − (5+x))/20] × 10

   Step 5 — Solve for x. 8.5 = [(25 − x)/20] × 10 = (25−x)/2 17 = 25 − x x = 8

   Step 6 — Use equation (1) to find y. y = 15 − x = 15 − 8 = 7

   Check. With x = 8, cf before median class = 5+8 = 13. Median = 20 + (30−13)/20 = 20 + 170/20 = 20 + 8.5 = 28.5 ✓. Total = 5+8+20+15+7+5 = 60 ✓

   ✦ Answer: x = 8, y = 7

   Where students slip. Solving only for x from the median equation and forgetting that y still needs the total-frequency equation to pin it down — the median condition alone never involves y at all, since y sits in a class after the median class.

   Another way. Notice that y doesn't appear anywhere in the median formula's working, because 40-50 comes after the median class — that's why the total-frequency equation is essential and not just a formality.

3. 35 marksNCERT Cl-10 Maths, Ex 13.3, Q3

   A life insurance agent found ages of 100 policy holders given as 'below' cumulative data: below 20 (2), below 25 (6), below 30 (24), below 35 (45), below 40 (78), below 45 (89), below 50 (92), below 55 (98), below 60 (100). Policies are given only from age 18 up to (but not including) 60. Calculate the median age.

   Hint. This is cumulative frequency data, not per-class frequency — recover each class's own frequency by subtracting consecutive cumulative values, and start the very first class at 18, not 0.

   Step 1 — Recover each class's own frequency by subtracting consecutive cumulative totals. Since policies start at 18, the first class is 18-20 with frequency 2 (matching 'below 20' directly). 18-20: 2 20-25: 6−2 = 4 25-30: 24−6 = 18 30-35: 45−24 = 21 35-40: 78−45 = 33 40-45: 89−78 = 11 45-50: 92−89 = 3 50-55: 98−92 = 6 55-60: 100−98 = 2

   Step 2 — Find n/2. n = 100, so n/2 = 50

   Step 3 — Build the cumulative frequency again from these classes to find the median class. 2, 6, 24, 45, 78, 89, 92, 98, 100 (this matches the original 'below' data, as it should) The first cumulative value reaching 50 is 78, in class 35-40.

   Step 4 — Read off the values and apply the formula. l = 35, cf = 45 (cumulative up to the previous class), f = 33, h = 5 Median = 35 + [(50−45)/33] × 5 = 35 + 25/33

   ✦ Answer: ≈35.76 years

   Where students slip. Starting the class intervals at 0 instead of 18. The question restricts policies to ages 18 and above, and the 'below 20' figure of 2 already reflects everyone from 18 up to 20 — treating the first class as 0-20 would badly misplace where the data actually starts.

   Another way. Since converting back to per-class frequencies reproduces the original cumulative numbers exactly (2, 6, 24, 45, 78, ...), that reconstruction is itself a built-in check that no subtraction was done incorrectly.

4. 44 marksNCERT Cl-10 Maths, Ex 13.3, Q4

   The lengths of 40 leaves, measured to the nearest millimetre: 118-126 (3), 127-135 (5), 136-144 (9), 145-153 (12), 154-162 (5), 163-171 (4), 172-180 (2). Find the median length.

   Hint. These classes have gaps (126 to 127) because the data was rounded to the nearest mm — the textbook's own hint says to convert to continuous classes first.

   Step 1 — Convert to continuous classes. Since measurements are rounded to the nearest mm, a length recorded as '126' could really be anywhere up to 126.5, and '127' could be as low as 126.5. Subtract 0.5 from every lower limit and add 0.5 to every upper limit: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5

   Step 2 — Build the cumulative frequency column. 3, 8, 17, 29, 34, 38, 40

   Step 3 — Find n/2. n = 40, so n/2 = 20

   Step 4 — Find the median class. 17 (up to 135.5-144.5) is short of 20; 29 (up to 144.5-153.5) reaches it. Median class is 144.5-153.5, with cf = 17.

   Step 5 — Apply the formula. l = 144.5, cf = 17, f = 12, h = 9 Median = 144.5 + [(20−17)/12] × 9 = 144.5 + 27/12 = 144.5 + 2.25

   ✦ Answer: 146.75 mm

   Where students slip. Applying the median formula directly to the printed classes (118-126, 127-135, ...) without the continuity correction. The formula assumes continuous classes with no gaps — skipping this step gives a class width and boundary that don't match reality.

   Another way. Once continuity-corrected, the class width becomes 9 throughout (126.5−117.5 = 9, and so on) — worth checking this comes out the same for every class, since an inconsistent width there would signal an error in the correction.

5. 54 marksNCERT Cl-10 Maths, Ex 13.3, Q5

   The distribution of the lifetimes of 400 neon lamps: 1500-2000 (14), 2000-2500 (56), 2500-3000 (60), 3000-3500 (86), 3500-4000 (74), 4000-4500 (62), 4500-5000 (48). Find the median lifetime of a lamp.

   Hint. Straightforward median procedure, just with larger numbers throughout.

   Step 1 — Set up the cumulative frequency column. 14, 70, 130, 216, 290, 352, 400

   Step 2 — Find n/2. n = 400, so n/2 = 200

   Step 3 — Find the median class. 130 (up to 2500-3000) is short of 200; 216 (up to 3000-3500) reaches it. Median class is 3000-3500, with cf = 130.

   Step 4 — Apply the formula. l = 3000, cf = 130, f = 86, h = 500 Median = 3000 + [(200−130)/86] × 500 = 3000 + [70/86] × 500

   Step 5 — Compute. ≈ 3000 + 406.98

   ✦ Answer: ≈3406.98 hours

   Where students slip. Using n (400) instead of n/2 (200) when comparing against the cumulative frequencies — the test is always against half the total, not the total itself.

   Another way. Since 216 is only just over 200, the median should sit close to the start of its class (l = 3000) rather than near its end — the computed 3406.98 is indeed in the lower half of the 3000-3500 range, consistent with that.

6. 65 marksNCERT Cl-10 Maths, Ex 13.3, Q6

   100 surnames were picked from a telephone directory and their letter counts distributed as: 1-4 (6), 4-7 (30), 7-10 (40), 10-13 (16), 13-16 (4), 16-19 (4). Determine the median, the mean, and the modal size of the surnames.

   Hint. All three again — this is good practice for keeping the three formulas straight in one sitting.

   Median first.

   Step 1 — Build the cumulative frequency column. 6, 36, 76, 92, 96, 100

   Step 2 — Find n/2. n = 100, so n/2 = 50

   Step 3 — Find the median class. 36 (up to 4-7) is short of 50; 76 (up to 7-10) reaches it. Median class is 7-10, with cf = 36.

   Step 4 — Apply the formula. l = 7, cf = 36, f = 40, h = 3 Median = 7 + [(50−36)/40] × 3 = 7 + 42/40 = 7 + 1.05

   ✦ Median = 8.05

   Mode.

   Step 5 — The modal class is 7-10 (highest frequency, 40). l = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3 Mode = 7 + [(40−30)/(2×40−30−16)] × 3 = 7 + [10/34] × 3 ≈ 7 + 0.882

   ✦ Mode ≈ 7.88

   Mean.

   Step 6 — Class marks: 2.5, 5.5, 8.5, 11.5, 14.5, 17.5. Assumed mean a = 8.5, h = 3: uᵢ: −2, −1, 0, 1, 2, 3 Σfᵢuᵢ = 6(−2)+30(−1)+40(0)+16(1)+4(2)+4(3) = −12−30+0+16+8+12 = −6 mean = 8.5 + 3(−6/100) = 8.5 − 0.18

   ✦ Mean = 8.32

   All three (8.05, 7.88, 8.32) sit close together, consistent with a fairly typical, moderately symmetric distribution of surname lengths.

   Where students slip. Reusing the median class's frequency and neighbours for the mode without checking they're actually the same class here — it happens that both median and mode fall in 7-10 for this dataset, but that has to be verified each time, not assumed.

   Another way. As a check on the mean: since the modal class (7-10) also carries the largest single frequency (40 out of 100), the mean should land reasonably close to that class's mark (8.5) — 8.32 does.

7. 74 marksNCERT Cl-10 Maths, Ex 13.3, Q7

   The weights of 30 students in a class: 40-45 (2), 45-50 (3), 50-55 (8), 55-60 (6), 60-65 (6), 65-70 (3), 70-75 (2). Find the median weight of the students.

   Hint. Standard procedure — build the cumulative frequency, find n/2, locate the class.

   Step 1 — Build the cumulative frequency column. 2, 5, 13, 19, 25, 28, 30

   Step 2 — Find n/2. n = 30, so n/2 = 15

   Step 3 — Find the median class. 13 (up to 50-55) is short of 15; 19 (up to 55-60) reaches it. Median class is 55-60, with cf = 13.

   Step 4 — Apply the formula. l = 55, cf = 13, f = 6, h = 5 Median = 55 + [(15−13)/6] × 5 = 55 + 10/6

   ✦ Answer: ≈56.67 kg

   Where students slip. Picking 50-55 as the median class because its cumulative frequency (13) is 'close to' 15. Close is not enough — the class must be the first one whose cumulative frequency actually reaches or exceeds n/2, and 13 falls short.

   Another way. With the modal class visibly at 50-55 (frequency 8, the highest) but the median class one step later at 55-60, this data has a slight rightward skew — worth noting as a quick check that median and mode need not coincide even in fairly regular-looking data.

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.