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Measures of Dispersion for Grouped Data · Form 5

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

Builds the basic skills for grouped data: finding class midpoints and the estimated mean Σfx ÷ Σf, naming the modal class, and locating the median class. Good for students starting this chapter who need each single step to feel automatic.

Written by the spmmath.com.my editorial team.

Worked example 1

The table shows the masses (kg) of 20 parcels. Estimate the mean mass.

Classes: 40–44 (4 parcels), 45–49 (6), 50–54 (7), 55–59 (3).

  1. Class midpoint x = (lower + upper) ÷ 2: 42, 47, 52, 57.
  2. fx for each class: 4×42=168, 6×47=282, 7×52=364, 3×57=171.
  3. Σf = 20 and Σfx = 168+282+364+171 = 985.
  4. Estimated mean = Σfx ÷ Σf = 985 ÷ 20 = 49.25.

Worked example 2

The table shows the time (minutes) 30 pupils took to finish a quiz: 0–9 (5 pupils), 10–19 (12), 20–29 (9), 30–39 (4). State the modal class.

  1. The modal class is the class with the highest frequency.
  2. Compare frequencies: 5, 12, 9, 4, the largest is 12.
  3. Frequency 12 belongs to the class 10–19.

Worked example 3

A group of 25 candidates scored as follows: 1–20 (3 candidates), 21–40 (7), 41–60 (10), 61–80 (5). Determine the class in which the median lies (the median class).

  1. Build cumulative frequency: 3, 3+7=10, 10+10=20, 20+5=25.
  2. Median position = n ÷ 2 = 25 ÷ 2 = 12.5th value.
  3. The cumulative frequency first reaches or passes 12.5 at 20, which is the class 41–60.

Worked example 4

The table shows the masses, in grams, of 25 mangoes: 100–119 (3), 120–139 (8), 140–159 (9), 160–179 (5). Find the range of the masses using the class boundaries.

  1. The lower class boundary of the first class (100–119) is 100 − 0.5 = 99.5 g.
  2. The upper class boundary of the last class (160–179) is 179 + 0.5 = 179.5 g.
  3. Range = upper boundary of the highest class − lower boundary of the lowest class.
  4. Range = 179.5 − 99.5 = 80 g.

Worked example 5

The table shows the times, in minutes, taken by 40 pupils to solve a puzzle: 5–9 (4), 10–14 (10), 15–19 (14), 20–24 (8), 25–29 (4). State the class in which the first quartile (Q₁) lies.

  1. Total frequency n = 4 + 10 + 14 + 8 + 4 = 40.
  2. The position of Q₁ is n/4 = 40/4 = the 10th value.
  3. Build cumulative frequencies: up to 9.5 → 4; up to 14.5 → 4 + 10 = 14.
  4. The 10th value falls between the 5th and 14th, which is in the class 10–14.

Worked example 6

The table shows the marks scored by 30 students in a test: 21–30 (4), 31–40 (7), 41–50 (10), 51–60 (6), 61–70 (3). Estimate the mean mark.

  1. Find the midpoint of each class: 25.5, 35.5, 45.5, 55.5, 65.5.
  2. Multiply each midpoint by its frequency (fx): 102, 248.5, 455, 333, 196.5.
  3. Sum: Σfx = 102 + 248.5 + 455 + 333 + 196.5 = 1335; Σf = 30.
  4. Mean = Σfx ÷ Σf = 1335 / 30 = 44.5.

Worked example 7

The table shows the number of minutes 24 students spent exercising daily: 10–19 (3), 20–29 (7), 30–39 (9), 40–49 (5). Estimate the mean number of minutes.

  1. Find the midpoint of each class: 14.5, 24.5, 34.5, 44.5.
  2. Find fx for each class: 3×14.5=43.5, 7×24.5=171.5, 9×34.5=310.5, 5×44.5=222.5.
  3. Σf = 3+7+9+5 = 24; Σfx = 43.5+171.5+310.5+222.5 = 748.
  4. Mean = Σfx/Σf = 748/24 ≈ 31.17 minutes.

Worked example 8

The table shows the heights, in cm, of 35 seedlings: 10–14 (4), 15–19 (9), 20–24 (12), 25–29 (7), 30–34 (3). Determine the class in which the third quartile (Q₃) lies.

  1. Find 3n/4 = 3(35)/4 = 26.25.
  2. Build the cumulative frequency: 10–14→4, 15–19→13, 20–24→25, 25–29→32, 30–34→35.
  3. Since 26.25 lies just after the cumulative frequency 25 (end of the 20–24 class), Q₃ falls in the next class.

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

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Frequently asked questions

What should I focus on first for easy grouped-dispersion questions?

Get comfortable reading a grouped frequency table correctly: identifying the class interval, calculating the class midpoint as (lower + upper limit) ÷ 2, and spotting the modal class as the interval with the highest frequency. Most easy questions test whether you can extract these values accurately before any calculation begins.

What earns marks on these easier grouped-data questions?

Correct midpoint values, a properly labelled table or histogram with equal class widths and no gaps between bars, and using the exact boundaries (not the stated limits) when class widths are calculated. Even a simple identification question needs the working shown to secure the mark, not just the final value.

What mistakes are common at this easy level?

Using the stated class limits instead of the class boundaries when finding class width, drawing histogram bars with gaps or uneven widths, and picking the class with the highest midpoint instead of the highest frequency as the modal class. Small table-reading slips here carry into every later calculation.

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