Measures of Dispersion for Grouped Data · 7.2.4
Solving problems on grouped dispersion
Students apply the range, interquartile range, variance and standard deviation to multi-step, real-life grouped-data problems, for example finding a missing frequency from a given mean, then using it to find the variance and standard deviation, combining earlier skills such as solving equations into one connected solution.
The official learning standard (7.2.4)
“Solve problems involving measures of dispersion for grouped data.”
What it means
Students apply the range, interquartile range, variance and standard deviation to multi-step, real-life grouped-data problems, for example finding a missing frequency from a given mean, then using it to find the variance and standard deviation, combining earlier skills such as solving equations into one connected solution.
How it is examined
Mainly Paper 2, as an extended question built on a grouped frequency table with one piece of information missing, often a frequency or the mean, requiring students to set up and solve an equation before finding the dispersion measure asked for. Paper 1 may test a single-step application, such as finding the range from given boundaries.
Worked example
The table shows the scores of a group of students in a competition, with one frequency unknown: 1–10 (4), 11–20 (k), 21–30 (6), 31–40 (2). Given that the mean score is 18.5, find the value of k, then determine the variance and standard deviation of the scores.
- Midpoints: 5.5, 15.5, 25.5, 35.5. Total frequency Σf = 4 + k + 6 + 2 = 12 + k.
- Σfx = 4(5.5) + k(15.5) + 6(25.5) + 2(35.5) = 22 + 15.5k + 153 + 71 = 246 + 15.5k.
- Set up the mean equation: (246 + 15.5k)/(12 + k) = 18.5 → 246 + 15.5k = 222 + 18.5k → 24 = 3k → k = 8.
- So Σf = 12 + 8 = 20 and Σfx = 246 + 15.5(8) = 370, giving x̄ = 370/20 = 18.5, matching the given mean.
- Σfx² = 4(5.5²) + 8(15.5²) + 6(25.5²) + 2(35.5²) = 121 + 1922 + 3901.5 + 2520.5 = 8465.
- Variance: σ² = Σfx²/Σf − x̄² = 8465/20 − 18.5² = 423.25 − 342.25 = 81.
- Standard deviation: σ = √81 = 9.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
How do I set up the equation when a frequency is missing?
Write every frequency, including the unknown one (e.g. k), into the Σf and Σfx expressions in terms of that letter, then substitute the given mean into x̄ = Σfx/Σf and solve the resulting linear equation.
This is the same equation-solving skill from earlier chapters, just applied inside a statistics context.
Once I've found the missing frequency, do I need to recompute the mean from scratch?
You can, as a check, recomputing x̄ = Σfx/Σf with the found frequency should give back the mean stated in the question. If it does not match, you have made an error solving for the unknown frequency, and it is worth reworking that step before continuing to the variance.
What other information might be missing in this type of question, besides a frequency?
A missing class boundary, a missing midpoint, or even working backward from a given standard deviation to find a frequency are also common. The method stays the same each time: write the unknown as a letter, form an equation from the given statistic, and solve.