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2 Responses
Think of an unequal-width histogram like a garden of flowerbeds: the width is how wide the bed is, the height is how densely it’s planted, and the area is how many flowers are actually in there. The rule that keeps everything sane is frequency density = frequency ÷ class width, and the bar’s area (height × width) represents the number of values in that interval. So: if they ask “which interval has the most values,” you look at area (equivalently, the frequency), not just the height-your 18/6 = 3 and 12/4 = 3 have equal heights (same density), but the first has more values because its area is bigger (18 vs 12). If they ask “where is the peak” or “modal class” in a histogram with unequal widths, that usually means the class with the highest frequency density-i.e., the tallest bar-because “mode” is about where the data are most concentrated per unit on the x-axis. A skinny, very tall bar can be the peak (highest density) even if it doesn’t contain the most values overall. Tiny caveat: this all assumes the vertical axis is frequency density (as it should be for unequal widths); if your plot literally labels the y-axis “frequency” with changing widths, that’s nonstandard and comparisons get messy. But in almost all test settings: most values → compare areas; peak/modal class → compare heights (densities).
You’ve got it: with unequal widths, the area of a bar tells you how many values are in that interval, but the “peak” is where the height (frequency density) is greatest. Think of each bar like a garden bed-its footprint (area) tells how many plants, while the tallest foliage (height) shows where they’re packed most tightly.