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Measures of Dispersion for Grouped Data · Form 5

Measures of Dispersion for Grouped Data: Paper 2 Answering Guide

How Measures of Dispersion for Grouped Data appears in Mathematics Paper 2 (Subjective), and how to lay out your working so you earn every method mark.

Written by the spmmath.com.my editorial team.

How it is examined

Paper 2 gives a frequency table and asks for the mean and standard deviation, or provides data to plot an ogive and then read the median and interquartile range from it. A carefully built table and a smooth, accurate curve are where the marks are.

Showing your working

Mathematics Paper 2 (Subjective) is worth 100 marks and gives marks for the steps, not only the answer. Write each line clearly: state the formula or rule, substitute the numbers, then simplify.

If the question carries units, carry them through to the final line.

Worked Paper 2 example: mean and standard deviation from a table

The table shows the heights, in cm, of 30 pupils. The class 150 ≤ h < 160 has 4 pupils, 160 ≤ h < 170 has 9 pupils, 170 ≤ h < 180 has 12 pupils, and 180 ≤ h < 190 has 5 pupils.

Calculate (a) the mean height, and (b) the standard deviation of the heights.

  1. State the rule and list the midpoints: for grouped data use the class midpoint x = (lower + upper) ÷ 2, giving x = 155, 165, 175, 185.
  2. Build the fx column: fx = 4(155), 9(165), 12(175), 5(185) = 620, 1485, 2100, 925; so Σf = 30 and Σfx = 5130.
  3. (a) Mean: x̄ = Σfx ÷ Σf = 5130 ÷ 30 = 171 cm.
  4. Build the fx² column: fx² = 4(155²), 9(165²), 12(175²), 5(185²) = 96100, 245025, 367500, 171125; so Σfx² = 879750.
  5. State the variance formula and substitute: σ² = Σfx² ÷ Σf − x̄² = 879750 ÷ 30 − 171² = 29325 − 29241 = 84.
  6. (b) Standard deviation: σ = √84 = 9.17 cm (3 s.f.).

The same table, a median-and-quartiles question

If the same data instead asks for the median and interquartile range using the ogive (or by interpolation), the cumulative frequencies 4, 13, 25, 30 drive the working.

  1. Median position n/2 = 30/2 = 15, which lands in the class 170–180 (the cumulative frequency passes 15 there): median = 170 + [(15 − 13)/12] × 10 = 171.7 cm.
  2. Lower quartile position n/4 = 7.5, in the class 160–170: Q1 = 160 + [(7.5 − 4)/9] × 10 = 163.9 cm.
  3. Upper quartile position 3n/4 = 22.5, in the class 170–180: Q3 = 170 + [(22.5 − 13)/12] × 10 = 177.9 cm.
  4. Interquartile range = Q3 − Q1 = 177.9 − 163.9 = 14.0 cm.

Source:SPM: Format Pentaksiran mulai 2021, Matematik (1449)

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Frequently asked questions

How many decimal places or significant figures should I give the standard deviation?

Follow the instruction in the question, commonly 'correct to 2 decimal places' or '3 significant figures'. If none is stated, give 3 or 4 significant figures and keep it consistent.

Round only at the very end; for √84 that gives 9.17 to 3 significant figures. Never round the mean or the totals first.

Do I lose marks if I skip the fx² column and just write the final answer?

Yes. Paper 2 awards method marks for the working, so a correct answer with no fx or fx² column and no formula line usually earns only the final mark.

If a single figure is wrong, you then score almost nothing. Showing the table and the substituted formula protects most of the marks.

What if the class widths are unequal, does this mean method still work?

Yes, for the mean and standard deviation. You still take each class's own midpoint, multiply by its frequency, and total the fx and fx² columns; unequal widths do not change that.

Unequal widths only matter for a histogram, where you would use frequency density rather than raw frequency for the bar heights.

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