Statistical Diagrams (AI HL)
Once a data set has been collected it's usually presented as a frequency table before anything else is calculated from it, and a surprising number of exam marks come from reading that table correctly. This page focuses on working with frequency tables - especially finding a mean or an unknown frequency - with worked examples and the mistakes that lose marks. It's part of the broader Statistics & Sampling topic.
11 questions on this sub-topic.
Presentation of discrete and continuous data
Covered under IB syllabus reference SL4.2: presentation of discrete and continuous data (frequency distributions, histograms with equal class intervals), plus the cumulative frequency, box-and-whisker and related diagrams built from a frequency table.
Mean from a frequency table
\(\bar{x} = \dfrac{\sum fx}{\sum f}\)
Multiply each value by its frequency, sum the products, then divide by the total frequency. This is standard technique rather than a formula-booklet entry - the GDC's one-variable statistics mode does it automatically once the values are entered as a frequency list.
Weighted mean
\(\bar{x} = \sum w_i x_i,\ \text{where}\ \sum w_i = 1\)
The same idea applied when categories carry unequal weightings (e.g. an exam worth 30% and a coursework worth 70%) rather than raw frequencies.
Need the full syllabus wording and formula-booklet reference table? See Statistics & Sampling.
Worked examples
A student scores 70 on a test weighted 30% and 85 on an exam weighted 70%.
Find the weighted mean.
Worked solution
Multiply each score by its weight: \(0.30(70)+0.70(85).\) M1
\(=21+59.5=80.5.\) A1
| Value | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Frequency | 3 | 5 | f | 2 |
The mean is 5. Find \(f\).
Worked solution
\(\dfrac{42+6f}{10+f}=5.\) M1
A1
\(42+6f=50+5f.\) M1
\(f=8.\) A1
For 200 exam scripts, the cumulative frequency table is shown below.
| Mark | \(\le40\) | \(\le50\) | \(\le60\) | \(\le70\) | \(\le80\) |
|---|---|---|---|---|---|
| Cumulative frequency | 30 | 80 | 150 | 185 | 200 |
(a) Estimate the 90th percentile.
(b) A grade A is awarded to the top 10%. State the minimum mark for an A (to nearest mark).
Worked solution
(a) 90th percentile is the \(0.9 \times 200 = 180\)th value, in \(60\!-\!70\): M1
\(60 + \dfrac{180 - 150}{35}\times10\) A1
\(\approx 68.6.\) A1
(b) Top 10% means at or above the 90th percentile, so the minimum A mark is about 69. M1 A1
Common mistakes
- Dividing by the number of distinct values instead of the total frequency. In a frequency table \(\sum f\) is the total number of data items, not the number of rows - dividing by 4 instead of \(3+5+f+2\) is a very common slip.
- Forgetting to multiply by \(x\) before summing. \(\sum fx\) means each value is multiplied by its own frequency first; adding the frequencies and values separately gives the wrong numerator entirely.
- Weights that don't sum to 1. In a weighted-mean question, always check the given percentages or fractions add to 100% (or 1) before using them - a typo in the question or a misread weight throws off the whole answer.
Ready to practise properly?
12 statistical-diagrams questions, marked instantly like the real exam.
Quick answers
How do you find the mean from a frequency table?
Multiply each value by its frequency, add these products together, and divide by the total frequency: \(\bar{x} = \dfrac{\sum fx}{\sum f}\).
How do you find a missing frequency if you're given the mean?
Write the mean formula with the unknown frequency as a letter, set it equal to the given mean, then cross-multiply and solve the resulting linear equation.