I’m revising stats and keep tripping over “averages from tables.” If I have a grouped frequency table (like ranges 0–10, 10–20, etc.), and the question says “find the average,” am I supposed to assume they mean the mean and use the midpoints for each class? Why is the midpoint the right thing to use, and does it still work if the class widths aren’t equal? I keep getting slightly different results if I try lower/upper bounds instead, which makes my brain do cartwheels.
Also, for the median from a grouped table: do I just find the median class with cumulative frequencies and then interpolate, or is taking the midpoint of the median class ever acceptable? I’m never sure when interpolation is expected versus “good enough.” Follow-up: what happens with edge cases like a final class that’s open-ended (e.g., 70+)? How do you pick a midpoint for that, or is there a standard workaround?
One more tiny thing that bugs me: if the classes are written as 0–10, 10–20, does it matter which end is inclusive when estimating the mean/median? Do I need to worry about that, or is it baked into the method?
I’d love a clear way to decide which approach to use so I stop second-guessing myself.
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3 Responses
Short answer: yes-estimate the mean with class midpoints (assumes values are uniformly spread within each class, so it’s fine with unequal widths), not the bounds; inclusivity at the edges doesn’t matter, and for a 70+ class you can’t get a reliable mean without assuming a width. For the median, find the median class with cumulative frequencies and linearly interpolate (the midpoint of that class is only a rough shortcut; if the median falls in 70+ you’d also need an assumed width): e.g., 0–10:2, 10–20:4, 20–30:1 gives mean ≈ (5·2+15·4+25·1)/7 = 13.57 and median ≈ 10 + ((4−2)/4)·10 = 15; nice walkthrough: https://www.mathsisfun.com/data/grouped-data.html
Use the midpoint for each class and estimate the mean as (sum of f×midpoint)/N under the “uniform within class” idea-this works with unequal widths, while using bounds introduces bias. For the median, identify the median class and interpolate L + ((N/2 − cumulative before)/f_in_class)×width (midpoint is only a rough fallback; open-ended classes prevent interpolation unless you assume a width), and inclusive endpoints are handled by taking consecutive half‑open intervals so you don’t need to worry.
For grouped data, the standard estimate of the mean is the weighted average of the class midpoints, using the class frequencies as weights. This implicitly assumes values are spread uniformly within each class, in which case the midpoint equals the class’s conditional mean; using a lower or upper bound instead would pretend every value sits at one edge and will usually bias the result. Unequal class widths do not change this: you still use midpoints with the actual frequencies; width only matters if you are given frequency densities, in which case first recover frequencies as density times width. For the median, find the class where the cumulative frequency first reaches half the total, then linearly interpolate within that class using its lower boundary, its width, and the fraction of the class needed to reach half the total; this is normally what examiners expect. Taking the midpoint of the median class is a rougher estimate and only “good enough” if the question invites approximation or the class is plausibly symmetric and narrow. Open-ended classes are awkward: you generally cannot estimate the mean without an extra assumption about tail behavior; for the median you are fine unless the median itself lies in an open-ended class, in which case interpolation needs an assumed width and the result is not reliable. Finally, for intervals like 0–10, 10–20, treat them as contiguous non-overlapping classes (e.g., 0 ≤ x < 10, 10 ≤ x < 20); the midpoint and the interpolation procedure don’t depend on which endpoint is written as inclusive, so you don’t need to worry about that. I’d use these conventions unless the question states otherwise.