Data summaries, histograms and interpretation
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| frequency density/ˈfriːkwənsi ˈdensɪti/ | mật độ tần số |
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?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The total is 25 across five observations: mean=25/5=5.
With unequal class widths the height is frequency density.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The sorted list is 2,4,4,6,9. Its third observation is 4.
Match each statistic to its definition.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.