Group continuous length records into intervals without losing boundaries or frequencies
Every valid measurement belongs to exactly one stated interval.
Core and Extended; Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
Before you begin
- Recall genes, alleles, genotype and phenotype from the inheritance pathway. Use the stated organism, units and sample; the records in these lessons are invented.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Where does exactly 50 mm belong under these interval rules?
What you will learn
- Construct non-overlapping measurement intervals
- Interpret a frequency graph with the original units
Original teaching model. Measurements, categories, DNA fragments and comparison records are invented; assumptions and sample boundaries are stated.
| Feature | Meaning |
|---|---|
| Classes | 40–<50 / 50–<60 / 60–<70 / 70–<80 mm |
| Counts | 4 / 7 / 6 / 3; total 20 |
| Percentages | 20% / 35% / 30% / 15% |
| Raw-data mean | 1168 mm ÷ 20 = 58.4 mm |
Construct non-overlapping measurement intervals
Grouping continuous measurements makes a larger record easier to describe, but the interval rule must be explicit. The original lengths use four ten-mm classes: 40 ≤ length < 50, 50 ≤ length < 60, 60 ≤ length < 70 and 70 ≤ length < 80. Each lower boundary is included and each upper boundary is excluded. A length of exactly 50 mm belongs in the second class, not both the first and second.
Count the twenty records once each. The four frequencies are 4, 7, 6 and 3, summing to twenty. Their percentages are twenty, thirty-five, thirty and fifteen per cent. The most frequent interval is 50 ≤ length < 60 mm with seven records. This does not mean every length in that class appeared, or that 55 mm is the most frequent exact observation. Class counts group observations and lose some detail.
Interpret a frequency graph with the original units
The histogram uses a continuous numerical axis and adjoining intervals. Because all four class widths are equal, frequency can be represented by bar height in this example. The vertical scale counts pods, not mm. Do not generalise the height-only rule to unequal-width classes without an appropriate density method; this lesson uses equal widths and does not require that extension.
Read the original records when calculating the exact mean or range. Their sum is 1168 mm, giving a mean of 58.4 mm for twenty pods, and the observed extremes give a 34 mm range. Interval midpoints would provide an estimate rather than the exact original mean. The chart describes an invented sample under its measurement protocol; it does not identify a gene, environmental cause, species-wide distribution or practical grade.
Pause and explain
What does the frequency 7 in the 50–<60 mm class show?
Worked example
Place 50 mm and 60 mm in the stated classes and check the total frequency.
Show the worked solution
- Use lower-inclusive and upper-exclusive boundaries: 50 mm is in 50–<60 and 60 mm in 60–<70.
- Add frequencies 4 + 7 + 6 + 3 = 20 and compare with the original specimen count.
Answer The boundaries place each value once; the frequencies account for all twenty valid specimens.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Place 49, 50, 59 and 60 mm in the stated ten-mm intervals.
Give me a hint
Apply the same lower-inclusive and upper-exclusive rule.
Compare my reasoning
- 49 belongs to 40–<50; 50 and 59 belong to 50–<60.
- 60 belongs to 60–<70.
49: first class; 50 and 59: second class; 60: third class.
Look for these in your work
- Correct boundaries
- No duplicate counts
Convert the four frequencies 4, 7, 6 and 3 out of 20 to percentages and identify the modal interval.
Give me a hint
The denominator counts valid specimens.
Compare my reasoning
- Divide each by 20 and multiply by 100.
- The largest count is 7 in 50–<60 mm.
Percentages 20%, 35%, 30%, 15%; the modal interval is 50 ≤ length < 60 mm.
Look for these in your work
- Correct percentages and total
- Modal interval with units
The original twenty length records sum to 1168 mm. Find the exact sample mean and explain why four midpoint values are insufficient for that exact mean.
Give me a hint
Use the raw-record sum and the twenty-specimen denominator.
Compare my reasoning
- 1168/20 = 58.4 mm.
- Class midpoints replace individual measurements and therefore produce an estimate.
The exact mean of the supplied records is 58.4 mm; a midpoint calculation would estimate it and does not recover all original values.
Look for these in your work
- 58.4 mm
- Exact versus grouped estimate
Common mistakes
- Including a boundary in two classes
- Reading a class frequency as an exact-value frequency
- Treating a midpoint mean as the exact original mean
What can you explain now?
What does the frequency 7 in the 50–<60 mm class show?
Compare with the explanation
Seven records lie in that interval
It groups several possible measured lengths.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Without your notes, construct non-overlapping measurement intervals and interpret a frequency graph with the original units. Keep units, the stated sample and the limits of the evidence clear.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.