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Measures of Dispersion for Grouped Data · Form 5

Measures of Dispersion for Grouped Data: Worked Examples (Medium)

Combines two or three steps: estimating the median and the quartiles by interpolation, finding the interquartile range, and back-solving for a missing frequency from a given mean. For students who can do the single steps and now need to link them.

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Worked example 1

The masses (g) of 40 eggs are grouped as: 20–29 (4), 30–39 (8), 40–49 (12), 50–59 (9), 60–69 (7). Estimate the median by interpolation.

  1. Cumulative frequency: 4, 12, 24, 33, 40, so Σf = n = 40.
  2. n ÷ 2 = 20; the cumulative frequency reaches 20 in class 40–49, so this is the median class.
  3. Read L = 39.5 (lower boundary), F = 12 (cumulative before), fm = 12, class size c = 10.
  4. Median = L + ((n/2 − F) ÷ fm) × c = 39.5 + ((20 − 12) ÷ 12) × 10.
  5. = 39.5 + (8 ÷ 12) × 10 = 39.5 + 6.67 = 46.17.

Worked example 2

The times (seconds) taken by 30 runners are grouped as: 1–10 (2), 11–20 (5), 21–30 (9), 31–40 (8), 41–50 (6). Find the interquartile range by interpolation.

  1. Cumulative frequency: 2, 7, 16, 24, 30; n = 30.
  2. Q1 position = n ÷ 4 = 7.5; it falls in class 21–30 (L = 20.5, F = 7, f = 9, c = 10).
  3. Q1 = 20.5 + ((7.5 − 7) ÷ 9) × 10 = 20.5 + 0.56 = 21.06.
  4. Q3 position = 3n ÷ 4 = 22.5; it falls in class 31–40 (L = 30.5, F = 16, f = 8, c = 10).
  5. Q3 = 30.5 + ((22.5 − 16) ÷ 8) × 10 = 30.5 + 8.13 = 38.63.
  6. Interquartile range = Q3 − Q1 = 38.63 − 21.06 = 17.57.

Worked example 3

The estimated mean of the data below is 33. Find the value of k.

Classes 10–19 (5), 20–29 (8), 30–39 (k), 40–49 (6), 50–59 (3).

  1. Midpoints x: 14.5, 24.5, 34.5, 44.5, 54.5.
  2. Σf = 5 + 8 + k + 6 + 3 = 22 + k.
  3. Σfx = 72.5 + 196 + 34.5k + 267 + 163.5 = 699 + 34.5k.
  4. Set mean = Σfx ÷ Σf = 33: (699 + 34.5k) ÷ (22 + k) = 33.
  5. 699 + 34.5k = 726 + 33k → 1.5k = 27 → k = 18.

Worked example 4

The table shows the number of hours 20 students spent on revision in a week: 0–2 (2), 3–5 (5), 6–8 (8), 9–11 (5). Estimate the mean and the standard deviation of the number of hours.

  1. Midpoints x: 1, 4, 7, 10. Frequencies f: 2, 5, 8, 5 (Σf = 20).
  2. Σfx = 2(1) + 5(4) + 8(7) + 5(10) = 2 + 20 + 56 + 50 = 128, so mean = 128/20 = 6.4 hours.
  3. Σfx² = 2(1²) + 5(4²) + 8(7²) + 5(10²) = 2 + 80 + 392 + 500 = 974.
  4. Variance = Σfx²/Σf − (mean)² = 974/20 − 6.4² = 48.7 − 40.96 = 7.74; standard deviation = √7.74 = 2.78 hours.

Worked example 5

The table shows the masses, in kg, of 50 chickens: 1.0–1.4 (6), 1.5–1.9 (12), 2.0–2.4 (18), 2.5–2.9 (9), 3.0–3.4 (5). Find the third quartile (Q₃) by interpolation.

  1. The position of Q₃ is 3n/4 = 3(50)/4 = 37.5th value.
  2. Cumulative frequencies: 6, 18, 36, 45, 50; the 37.5th value lies in the class 2.5–2.9 (cf just below = 36).
  3. For this class: lower boundary L = 2.45, cf before F = 36, frequency f = 9, class width c = 0.5.
  4. Q₃ = L + ((3n/4 − F)/f) × c = 2.45 + ((37.5 − 36)/9) × 0.5 = 2.45 + 0.083 = 2.53 kg.

Worked example 6

The table shows the number of units sold per day at a stall over 40 days: 10–19 (5), 20–29 (8), 30–39 (15), 40–49 (7), 50–59 (5). Estimate the mode using the modal-class formula.

  1. The modal class is 30–39 because it has the highest frequency (15).
  2. Lower boundary L = 29.5; Δ₁ = 15 − 8 = 7 (against the class before); Δ₂ = 15 − 7 = 8 (against the class after); class width c = 10.
  3. Mode = L + (Δ₁/(Δ₁ + Δ₂)) × c.
  4. Mode = 29.5 + (7/(7 + 8)) × 10 = 29.5 + (7/15) × 10 = 29.5 + 4.67 = 34.17 units.

Worked example 7

The table shows the masses, in kg, of 45 fish caught in a competition: 0.5–0.9 (5), 1.0–1.4 (10), 1.5–1.9 (15), 2.0–2.4 (10), 2.5–2.9 (5). Estimate the mean mass and the standard deviation of the masses.

  1. Midpoints: 0.7, 1.2, 1.7, 2.2, 2.7. Σf = 45.
  2. Σfx = 5(0.7)+10(1.2)+15(1.7)+10(2.2)+5(2.7) = 3.5+12+25.5+22+13.5 = 76.5. Mean = 76.5/45 = 1.7 kg.
  3. Σfx² = 5(0.49)+10(1.44)+15(2.89)+10(4.84)+5(7.29) = 2.45+14.4+43.35+48.4+36.45 = 145.05.
  4. Variance = Σfx²/Σf − mean² = 145.05/45 − 1.7² = 3.2233 − 2.89 = 0.3333.
  5. Standard deviation = √0.3333 ≈ 0.577 kg.

Worked example 8

The table shows the time, in minutes, spent by shoppers in a store: 10–19 (6), 20–29 (m), 30–39 (14), 40–49 (5). If the median time is 27 minutes, find the value of m.

  1. Total frequency n = 6+m+14+5 = 25+m.
  2. The median class is 20–29 (lower boundary L = 19.5, frequency f = m, cumulative frequency before it F = 6, class width c = 10).
  3. Apply the median formula: 27 = 19.5 + [(n/2 − F)/f] × c = 19.5 + [((25+m)/2 − 6)/m] × 10.
  4. 7.5 = [((25+m)/2 − 6)/m] × 10 → 0.75m = (25+m)/2 − 6 = 12.5 + 0.5m − 6.
  5. 0.75m − 0.5m = 6.5 → 0.25m = 6.5 → m = 26.

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What's new at medium level compared to the easy grouped-dispersion questions?

Medium questions usually ask you to build a cumulative frequency table and draw an ogive, then read the median and quartiles directly off the curve, or to calculate the range and interquartile range from a grouped table using upper and lower boundaries. You're now working with the whole dataset, not just single-class values.

What do examiners want to see on ogive and quartile questions?

A smooth curve through the upper boundary of each class plotted against cumulative frequency, correctly scaled axes, and clearly marked construction lines showing where you read off the median or quartile values. The reading itself must match your curve, an answer that doesn't align with the drawn ogive loses marks.

What common errors show up at this medium level?

Plotting cumulative frequency against the class midpoint instead of the upper boundary, forgetting to start the ogive from zero at the lower boundary of the first class, and misreading N/4, N/2 or 3N/4 positions on the vertical axis when locating quartiles. These small setup errors shift every value read off the curve.

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