Measures of Central Tendency (AI SL)
Mean, median and mode all answer the same question - "what's a typical value?" - but they answer it in different ways, and IB questions expect you to know which one a real situation calls for. This page covers all three, including the weighted mean for frequency tables and the estimated mean for grouped data, with worked examples and the mistakes that lose marks. It's part of the broader Descriptive Statistics topic.
40 questions on this sub-topic.
Mean, weighted mean and estimated mean
Covered under IB syllabus reference SL4.3: measures of central tendency (mean, median, mode), including estimating the mean from grouped data.
Mean from a frequency table
\(\bar x = \dfrac{\sum fx}{\sum f}\)
In the formula booklet (for equal class intervals). Each value \(x\) is weighted by how often it occurs, \(f\); don't just average the distinct values.
Estimated mean from grouped data
\(\bar x \approx \dfrac{\sum fx}{\sum f}\), \(x = \) class midpoint
Same formula from the formula booklet, applied with midpoints in place of exact \(x\)-values. Since exact values inside a class are unknown, the midpoint stands in for every value in that class - hence "estimate".
Median and mode are read directly from the data, not calculated from a formula. Need the full syllabus wording? See Descriptive Statistics.
Worked examples
Scores: 2, 2, 2, 3, 3, 3, 3, 3, 4, 4.
Find the mean.
Worked solution
With a frequency table each value \(x\) is repeated \(f\) times, so \(\bar x=\dfrac{\sum fx}{\sum f}\) (not the plain average of the distinct values). M1
\(2(3)+3(5)+4(2)=6+15+8=29.\) A1
\(\bar x=\dfrac{29}{10}=2.9.\) A1
Times (minutes) are grouped as follows:
| Class | 0–10 | 10–20 | 20–30 |
|---|---|---|---|
| Frequency | 4 | 9 | 7 |
(a)(i) State the midpoint of \(0\text{–}10.\)
(a)(ii) State the midpoint of \(10\text{–}20.\)
(a)(iii) State the midpoint of \(20\text{–}30.\)
(b) Estimate the mean.
Worked solution
(a)(i) Midpoints. For grouped data we assume every value sits at the centre of its class: midpoint of 0–10 \(=\tfrac{0+10}{2}=5.\)
(a)(ii) Midpoint of 10–20 \(=\tfrac{10+20}{2}=15.\)
(a)(iii) Midpoint of 20–30 \(=\tfrac{20+30}{2}=25.\) A1
(b) Use the midpoints as the \(x\)-values in the weighted mean:
\(\bar x=\dfrac{\sum fx}{\sum f}=\dfrac{5(4)+15(9)+25(7)}{4+9+7}=\dfrac{20+135+175}{20}=\dfrac{330}{20}=16.5\text{ minutes.}\) M1
\(\bar x=16.5\) minutes. A1 This is an estimate because the exact values inside each class are unknown.
Common mistakes
- Averaging the distinct values instead of using frequencies. Given a frequency table, the mean is \(\bar x=\dfrac{\sum fx}{\sum f}\), not the plain average of the values that appear - a value seen ten times must count ten times.
- Using the class boundary instead of the midpoint for an estimated mean. Grouped-data means always use the midpoint of each class as the representative \(x\)-value, never the upper or lower boundary.
- Missing a second (or third) mode. If more than one value shares the highest frequency, state every one of them - the data is bimodal or multimodal, and giving only one is an incomplete answer.
Ready to practise properly?
41 central-tendency questions, marked instantly like the real exam.
Quick answers
Why is the mean from grouped data called an estimate?
Grouped data only tells you how many values fall in each class, not their exact values, so the calculation uses the class midpoint as a stand-in for every value in that class - the result is close to, but not exactly, the true mean.
Can a data set have more than one mode?
Yes. If two values are tied for the highest frequency the data is bimodal, and both values are stated as modes; if three or more are tied it's multimodal. See using your GDC on the full topic page for entering frequency lists.