Measures of Dispersion for Grouped Data · Form 5
Measures of Dispersion for Grouped Data: Common Mistakes
The mistakes that quietly cost marks in Measures of Dispersion for Grouped Data, and how to avoid each one in the SPM exam.
In our experience teaching Measures of Dispersion for Grouped Data, most lost marks come from a small set of repeated slips, not from a lack of understanding. Here they are, with the fix for each.
Mistakes to avoid
- Using class boundaries when the midpoint is needed (or vice versa)
- Plotting an ogive against the wrong x-value
- Reading the median off a rushed, uneven curve
Seven more slips that quietly cost marks
- What students write: Σfx² = (Σfx)² = 5130². → Why it loses marks: it squares the grand total instead of summing each f × x², so the variance becomes meaningless. → Correct working: square each midpoint first, multiply by its frequency, then add: Σfx² = 4(155²) + 9(165²) + 12(175²) + 5(185²) = 879750.
- What students write: variance = 84, so standard deviation = 84. → Why it loses marks: the variance is in squared units; the standard deviation is its square root. → Correct working: standard deviation = √84 = 9.17.
- What students write: median position = (n + 1)/2 = 31/2 = 15.5. → Why it loses marks: (n + 1)/2 is the ungrouped rule; grouped data reads the ogive at n/2. → Correct working: use n/2 = 30/2 = 15 and read the median across from 15.
- What students write: modal class = 12. → Why it loses marks: 12 is a frequency, not a class, so it cannot be the modal class. → Correct working: the highest frequency, 12, belongs to the interval 170–180, so the modal class is 170–180.
- What students write: median = L + [(N/2 − F)/f] × c with F as the median class's own frequency. → Why it loses marks: F must be the cumulative frequency BEFORE the median class, not that class's frequency. → Correct working: with F = 13, median = 170 + [(15 − 13)/12] × 10 = 171.7.
- What students write: mean ≈ 171 ≈ 170, then variance = Σfx²/Σf − 170². → Why it loses marks: rounding the mean before squaring shifts the variance and the standard deviation. → Correct working: keep the exact mean 171, so variance = 29325 − 29241 = 84.
- What students write: interquartile range = Q1 − Q3. → Why it loses marks: the order is reversed, giving a negative spread. → Correct working: interquartile range = Q3 − Q1, always the upper quartile minus the lower.
A 30-second self-check before you move on
Before writing the final answer, run four quick checks: (1) is Σfx² built from f × x² and not (Σfx)²? (2) did you square-root the variance to get the standard deviation?
(3) is the modal class an interval, not a frequency? (4) is the interquartile range Q3 − Q1, a positive number?
Catching any one of these here saves the marks the seven slips above quietly take.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
My standard deviation came out larger than the mean, did I go wrong?
Not necessarily. The standard deviation and the mean measure different things, so there is no rule that one must be smaller.
For data with values near zero or a wide spread, the standard deviation can exceed the mean. Re-check your Σfx² and the square root, but a large value alone is not an error.
Can the variance be negative, or the number under the square root come out negative?
No. Variance is an average of squared deviations, so it is always zero or positive.
If Σfx²/Σf − x̄² gives a negative number, you have an arithmetic error, usually Σfx² too small or the mean squared too large. Recompute the fx² column before taking any root.
A friend used Σf(x − x̄)² and I used Σfx²/Σf − x̄². Why do we get the same answer?
Because they are the same formula rearranged. Σfx²/Σf − x̄² is the algebraic shortcut for the mean of the squared deviations, so both give an identical variance.
The shortcut is usually faster because it avoids subtracting the mean from every midpoint. Pick whichever the paper prints and stay consistent.