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7 Responses
Yep-use frequency density: set each bar’s height to frequency ÷ class width so the area equals the frequency (or use relative frequency ÷ width if you want areas to sum to 1). If the heights are ugly, scale them all by the same constant; the key is height × width matches the counts, pretty sure that’s what you need.
Exactly-use a density histogram: set each bar’s height to frequency ÷ class width so area tracks frequency (or frequency ÷ (total × width) if you want areas to represent proportions/probabilities); nice explainer here: https://www.bbc.co.uk/bitesize/guides/zw2hsrd/revision/3. Hope this helps!
Use frequency density: set bar height = frequency / class width so that area = frequency (and if you want probabilities instead, use frequency / (N × class width)). Think of pouring the same soup into wider bowls-the level (height) drops to keep the amount consistent; quick walkthrough here: https://www.bbc.co.uk/bitesize/guides/z6y9jty/revision/1.
Yes-use frequency density: set each bar’s height to frequency ÷ class width (or relative frequency ÷ width if you want total area 1), so that each bar’s area equals its frequency. Analogy: pour the same paint into trays of different widths-the wider the tray, the lower the level must be to represent the same amount.
Yep-use frequency density: set each bar’s height to frequency ÷ class width (or relative frequency ÷ width if you want total area 1), so height × width = frequency and the areas tell the true story. Think of pouring the same sand into trays of different widths-the narrower tray has to be taller to hold the same amount.
You’re on exactly the right track: when bin widths are unequal, set the bar height to the frequency density, i.e., height = frequency ÷ class width (in the same units as your data). That way, area = width × height = width × (frequency/width) = frequency, so the areas “tell the story.” If you want the histogram to represent probabilities instead, use relative frequency density: height = (frequency/n) ÷ width, which makes the total area 1. It’s also fine to multiply all heights by the same constant to get nicer axis numbers-just label the vertical axis “frequency density” (or note the scaling). Quick check: if all bins had the same width, this reduces to a regular histogram because the heights are then just proportional to the frequencies. Tiny example: a 0–5 bin with frequency 20 gets height 20/5 = 4; a narrower 5–7 bin with frequency 12 gets height 12/2 = 6-taller bar, same area logic. Analogy time: think of pouring the same amount of juice into a skinny glass versus a wide mug-the skinny one looks taller, but it’s the same “amount,” just spread differently. A nice recap with examples is here: https://www.bbc.co.uk/bitesize/guides/zcns7p3/revision/1
Love this question! When bins aren’t equal, the picture only tells the truth if area carries the count. So you set the bar height to frequency divided by class width. That way height × width = area = frequency. You were absolutely on the right track. I like to label the vertical axis “frequency density,” and if you want the areas to add up to 1 (so it feels like probabilities), divide by the total sample size as well. Quick feel check: if one bin has width 5 and frequency 20, its height is 20/5 = 4; if another has width 2 and the same frequency 20, its height is 20/2 = 10. Different heights, same area 20, so the story is fair.
Analogy time: imagine buckets of popcorn in different-sized buckets. Raw counts favor the bigger bucket, but “popcorn per inch of bucket” lets you compare how packed they really are-that’s frequency density. I’m always curious about this because it’s one of those places where pictures can quietly mislead us unless we nudge them into being fair. If the bin widths don’t vary too wildly, some people just plot raw frequencies and let the widths “sort it out,” and others rescale by the largest class width to keep the tallest bar visually familiar-those are more cosmetic than canonical, though. The core idea is still: divide by the class width so area matches the count.