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Cambridge IGCSE Mathematics — 0580 Extended

Topic 9.4: Statistics — Grouped Data

Once data are in classes you no longer know each value. The estimated mean treats every observation as the class midpoint. The modal class is the class with the largest frequency — not a single number.

Midpoints

Midpoint \(= \) (lower end \(+\) upper end) \(/ 2\). For \(20 \le t < 30\) that is 25. Open or closed ends do not change the midpoint on 0580.

Number line of class 20 to 30 with midpoint 25 marked
Every journey in this class is represented by 25 in the estimated mean.

Estimated mean

Add a midpoint column \(x\) and a product column \(fx\). Then

\[ \text{estimated mean} = \dfrac{\Sigma fx}{\Sigma f} \]

Method

  1. Write midpoints.
  2. Multiply \(f \times x\) in each row.
  3. Add the \(f\) column and the \(fx\) column.
  4. Divide. Round only at the end (3 s.f. unless the question says otherwise).

Estimate the mean journey time.

Frequency table with midpoints and fx, estimated mean 21.7 minutes
\(\Sigma fx = 650\), \(\Sigma f = 30\), so \(650/30 = 21.7\) min (3 s.f.).

Modal class

The class with the largest \(f\). Quote the interval, not its midpoint.

Equal-width frequency diagram with 20 to 30 highlighted as the modal class
Largest bar: \(20 \le t < 30\). Do not write ‘the mode is 25’.
Unknown true times in a class versus all stacked at the midpoint 25
If the times really cluster at 21, the estimate 25 is a little high. You cannot know.

Grouped discrete

Classes such as 1–3, 4–6 sit on integers. The midpoint of 1–3 is \((1+3)/2 = 2\).

Grouped discrete table of books with estimated mean 4.7
\(94/20 = 4.7\) books.

Paper 2 (non-calculator)

\(650 \div 30 = 21.\overline{6}\). The paper will accept \(21.7\) (3 s.f.) or \(65/3\) if it asks for a fraction.

Try this

In the journey table, which class is the modal class?

Show answer
Answer
  1. Largest frequency is 12.

    \[ 20 \le t < 30 \]

Exam Traps

  • Never call \(21.7\) “the mean”. It is an estimate of the mean — the papers award the word.

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