Chapter 14.4 — Exercise 14.3 — Median of Grouped Data

Two Quick Examples, Odd and Even

Adding a single new value (16, in the second list) doesn't just shift the median slightly — it can flip which of the two rules applies entirely, since it changes n from odd to even. Every grouped-data problem in this exercise builds toward the same underlying idea, just applied to data too large to sort and count by hand one observation at a time.

Cumulative Frequency: A Genuinely New Column

Grouped data can't be individually sorted and counted to the middle — there's no way to point at "the 34th house" inside a class interval covering dozens of houses at once. The workaround is a new column neither the mean nor the mode formulas ever needed: cumulative frequency (cf), a running total of fᵢ built up class by class from the very first row.

Locating the median class is really just the grouped-data version of finding the "middle position" from the ungrouped rule above — n/2 plays the same role that "4th value" or "average of 4th and 5th" played there, just located inside a running total instead of inside a sorted list.

Electricity Consumption: All Three Measures Together

Sixty-eight consumers' monthly electricity consumption, grouped into seven 20-unit classes, asks for the median, the mean, and the mode side by side — the fullest three-way comparison in the whole chapter.

All three land close together — median 137, mean ≈137, mode ≈135.8 — which is exactly the signature of a fairly evenly spread, only mildly skewed dataset. When mean, median, and mode all cluster this tightly, it's a strong visual and numeric signal that the underlying distribution is close to symmetric, without needing to plot the actual distribution to see it.

Solving for Two Missing Frequencies

A six-class table has two unknown frequencies, x and y, spread across two different classes — but the median (28.5) and total count (n=60) are both already given, supplying exactly enough information to solve for both unknowns together.

Notice which class the median 28.5 was declared to fall in before any algebra even started — that's the one genuinely new piece of information this problem supplies beyond the raw total. Without being told which class contains the median, x couldn't be isolated at all, since the median-class formula itself is what turns the word "28.5" into a solvable equation for x.

When Cumulative Frequency Is Given Directly

A hundred insurance policy holders' ages are given as a "less than" cumulative table from the start — "below 20," "20–25," and so on — rather than as raw class frequencies. The individual fᵢ values have to be recovered first, by subtracting each cumulative total from the one before it.

Every other problem in this exercise starts with fᵢ and builds cf up as a running total; this one runs that exact process backward, starting from cf and recovering fᵢ by subtraction. Once the individual frequencies are recovered, though, the median formula itself works identically to every other problem here — the cf column was needed anyway, so having it handed over directly actually saves a step rather than adding one.

Inclusive Classes Need Continuous Boundaries First

Forty leaves' lengths are grouped into classes written as "118–126," "127–135," and so on — gaps of 1 unit between consecutive classes (126 to 127) rather than classes that meet exactly. Before any median formula can apply, these inclusive classes need converting into continuous ones by extending each boundary by half a unit.

Skipping the boundary conversion and using the original "118, 127, 136…" limits directly would silently shift every single class mark and boundary by half a unit, throwing off l in the final formula by exactly that same half-unit — a small-looking adjustment that matters precisely because it's easy to skip without immediately noticing anything obviously wrong in the arithmetic that follows.

Three More, Compressed

The surnames problem is worth a second look for how close its three measures land — 8.05, 8.32, and 7.88, all within half a letter of each other — a tighter three-way cluster even than the electricity-consumption problem earlier, suggesting a genuinely well-balanced, close-to-symmetric spread in how many letters people's surnames tend to have.

Onward to the Graph

Every median in this exercise came from the same formula, l + [(n/2−cf)/f]×h, applied to a cumulative frequency column built up class by class. Exercise 14.4 takes that exact same cumulative frequency column and turns it into a graph — plotting it directly reveals the median as a specific point on a curve, reachable without touching the formula at all. Revisit Exercise 14.2 and Exercise 14.1 for the mode and mean formulas this exercise repeatedly compared its own results against.