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6.2.2Inquiry Process

Processing Data in Biology — IB Biology SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 7 of 10

📚 IB Biology SL Hub / Topical Revision
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Part 1 of 5Core Calculations

6.2.2 Processing Data in Biology

Processing turns raw measurements into values that can answer the research question.

Mean
  • Add the replicate values and divide by their number, leaving out any result that has been justified as anomalous.
Percentage change
  • Use (final value − initial value) ÷ initial value × 100, so samples with different starting sizes can be compared.
Rate from time
  • When the time for an event is measured, calculate rate as 1 ÷ time, which has units of s⁻¹.
Rate from change
  • When a quantity changes, rate = change in the quantity ÷ time taken, for example in cm³ s⁻¹.
Standard deviation
  • This measures the spread of values around the mean, and a small value shows the repeats were close together.

Part 2 of 5Presenting Processed Data

The route from raw to processed values must be clear to a reader.

1
Keep raw data in one table and show processed values in a second, separate table with its own title.
2
Show one complete sample calculation for each type of calculation, with the units written at every step.
3
Give each final answer to the same number of significant figures as the least precise data used.
4
Include a measure of spread, such as the standard deviation, beside every mean in the processed table.
5
Do not round intermediate numbers heavily, and round only when stating the final answer.

Part 3 of 5Using Spread and Uncertainty

Using Spread and Uncertainty
Small SD
  • The repeats are close together, so the measurements are precise.
Large SD
  • The values are widely scattered, often because of natural variation or random error in the method.
Sample SD
  • Spreadsheets and calculators give the sample standard deviation, which divides by n − 1 for a small set of repeats.
Significant figures
  • A calculator display with many digits suggests a precision that the original readings did not have.

Part 4 of 5Worked Example

Question

  • Three potato cylinders in 0.6 mol dm⁻³ sucrose have initial masses 1.82 g, 1.85 g and 1.79 g and final masses 1.65 g, 1.70 g and 1.63 g.
  • Find the mean percentage change in mass and the standard deviation.

  1. Cylinder 1: (1.65 − 1.82) ÷ 1.82 × 100 = −9.34 %.
  2. Cylinder 2: (1.70 − 1.85) ÷ 1.85 × 100 = −8.11 %.
  3. Cylinder 3: (1.63 − 1.79) ÷ 1.79 × 100 = −8.94 %.
  4. Mean = (−9.34 − 8.11 − 8.94) ÷ 3 = −8.80 %.
  5. Sample standard deviation (n − 1) = 0.63 %.

Answer Mean change in mass = −8.8 % with a standard deviation of 0.6 %.

Part 5 of 5Exam Checklist & Tips

6.2.2 – Key Points for the Exam

  • Percentage change The change in a quantity as a percentage of its starting value.
  • Rate How quickly a quantity changes, found from a change divided by time, or 1 ÷ time for an event.
  • Standard deviation A measure of how far values are spread around the mean.
  • Significant figures The meaningful digits of a number, limited by the precision of the raw data.
  • Sample calculation One full worked example showing how a processed value was obtained.

Common Mistakes

  • Giving all the digits on the calculator display instead of rounding to the right significant figures.
  • Averaging in a result that was already identified as anomalous without explaining why.
  • Showing only a spreadsheet output with no sample calculation to prove the method.
  • Using the absolute change in mass when the starting masses differ, instead of the percentage change.

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