Data summaries, histograms and interpretation
Введено| English | Значение |
|---|---|
| frequency density/ˈfriːkwənsi ˈdensɪti/ | плотность частоты |
Can one average tell the whole story?
- Two groups have the same median but different spread. One summary cannot describe both location and consistency.
- This lesson studies frequency density 频率密度: Frequency divided by class width, used as histogram height.
The method
- Compare an appropriate average and spread in context.
- A histogram uses area for frequency, so height=frequency/class width.
- Grouped estimates assume representative values within intervals.
In a histogram with unequal class widths, which bar contains the most observations?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Histogram area equals frequency. A wide bar of low density can hold more observations than a narrow tall bar; height alone shows density, not frequency.
A worked case
A class from 10 to 20 with frequency 30 has density 30/10=3. A class from 20 to 40 with frequency 20 has density 20/20=1. Its wider bar must not be mistaken for a larger density. For values 2,4,4,6,9, the median is 4 and mean is 5.
Data summaries, histograms and interpretation
Compare an appropriate average and spread in context
Choose a centre and a spread measure for the worked observations.
Find the density for frequency 30 and class width 10.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Density=frequency/class width=30/10=3.
Where it goes wrong
- The tallest histogram bar need not contain the most observations.
- A grouped mean is an estimate.
- Correlation does not prove causation, and extrapolation extends beyond the observed range.
Not true: A histogram bar's height equals its frequency, even with unequal class widths. With unequal class widths the height is frequency density; frequency is the area of the bar.
Find the mean of 2,4,4,6,9.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The total is 25 across five observations: mean=25/5=5.
With unequal class widths the height is frequency density.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. With unequal class widths the height is frequency density; frequency is the area of the bar.
Applying it
- Choose a display that fits the data type.
- Give both a numerical comparison and what it means for the population; do not infer more precision than the sample supports.
Find the median of 2,4,4,6,9.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The sorted list is 2,4,4,6,9. Its third observation is 4.
Match each statistic to its definition.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The mean uses every value, the median resists outliers, and frequency density makes histogram areas carry the frequencies.
In your course
- AQA A-level Mathematics (7357), specification section L.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Frequency density: Frequency divided by class width, used as histogram height.
- Compare an appropriate average and spread in context.
- Watch for: the tallest histogram bar need not contain the most observations.