Measures of Dispersion for Grouped Data · 7.2.1
Dispersion measures for grouped data
Students calculate four measures of spread for grouped data: the range (upper boundary of the highest class minus lower boundary of the lowest), the interquartile range (Q3 − Q1, from the ogive or by interpolation), the variance σ² = Σfx²/Σf − x̄², and the standard deviation σ = √σ², using the class midpoint x and frequency f from a frequency table.
The official learning standard (7.2.1)
“Determine the range, interquartile range, variance and standard deviation as measures to describe the dispersion of grouped data.”
What it means
Students calculate four measures of spread for grouped data: the range (upper boundary of the highest class minus lower boundary of the lowest), the interquartile range (Q3 − Q1, from the ogive or by interpolation), the variance σ² = Σfx²/Σf − x̄², and the standard deviation σ = √σ², using the class midpoint x and frequency f from a frequency table.
How it is examined
Mainly Paper 2: given a grouped frequency table, students build midpoint, fx and fx² columns to compute the mean, variance and standard deviation, and use a cumulative frequency column (or an ogive) for the quartiles and interquartile range. Paper 1 may test a single formula application, such as finding the range from given class boundaries.
Worked example
The table shows the heights (cm) of 30 plants in a nursery: 10–19 (3), 20–29 (7), 30–39 (12), 40–49 (6), 50–59 (2). Determine (a) the range, (b) the interquartile range, (c) the variance, and (d) the standard deviation of the heights.
- (a) Range = upper boundary of the highest class − lower boundary of the lowest class = 59.5 − 9.5 = 50 cm.
- (b) Cumulative frequencies: 3, 10, 22, 28, 30 (N = 30). Q1 at N/4 = 7.5 falls in the class 20–29 (boundaries 19.5–29.5): Q1 = 19.5 + [(7.5 − 3)/7] × 10 = 25.9.
- Q3 at 3N/4 = 22.5 falls in the class 40–49 (boundaries 39.5–49.5): Q3 = 39.5 + [(22.5 − 22)/6] × 10 = 40.3. Interquartile range = 40.3 − 25.9 = 14.4 cm.
- (c) Midpoints x = 14.5, 24.5, 34.5, 44.5, 54.5. Σfx = 1005, so x̄ = 1005/30 = 33.5.
- Σfx² = 630.75 + 4201.75 + 14283 + 11881.5 + 5940.5 = 36 937.5.
- σ² = Σfx²/Σf − x̄² = 36 937.5/30 − 33.5² = 1231.25 − 1122.25 = 109 cm².
- (d) σ = √109 ≈ 10.4 cm.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
Why use class boundaries instead of class limits for the range?
Grouped data hides the true highest and lowest individual values, so the range is estimated using the upper boundary of the last class minus the lower boundary of the first class. Boundaries, not limits, are used because they mark the actual continuous edges of the data, consistent with how the ogive and histogram are drawn.
Do I always need the mean before finding the variance?
Yes. Even the short formula σ² = Σfx²/Σf − x̄² needs x̄ = Σfx/Σf, so calculate the mean first, keep it unrounded on your calculator, then square it for the variance step.
Rounding the mean too early is a common source of an inaccurate final answer.
Which measure should I use if the data has an unusually high or low class?
The interquartile range is more reliable than the range in this situation, since it only covers the middle half of the data and ignores the extremes. The standard deviation is still useful and is usually preferred for general comparisons, but remember it is affected by every value, including unusually high or low ones.