Data summaries, histograms and interpretation · Higher
| English | 中文 | Pinyin |
|---|---|---|
| frequency density/ˈfriːkwənsi ˈdensɪti/ | 频率密度 | pín lǜ mì dù |
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.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines frequency density?
Frequency divided by class width, used as histogram height.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
Density=frequency/class width=30/10=3.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
A histogram bar's height always equals its frequency, even with unequal class widths. This claim is false. Explain which definition or assumption it violates.
Find the mean of 2,4,4,6,9.
The total is 25 across five observations: mean=25/5=5.
A histogram bar's height always equals its frequency, even with unequal class widths.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the median of 2,4,4,6,9.
The sorted list is 2,4,4,6,9. Its third observation is 4.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 8300 · Higher · 3.6. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
Frequency divided by class width, used as histogram height. Choose the relationship, show the method, check its assumptions and interpret the result.