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Estimated Mean

FoundationHigherAQAEdexcelOCR

Master estimated mean for GCSE Maths with structured, exam-style practice. This Foundation and Higher resource covers estimating the mean from grouped data and includes a complete mark scheme showing the steps examiners reward. Suitable for AQA, Edexcel and OCR. Free to download as a PDF. Use the midpoint of each class, since you don't have exact values.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

Foundation

WorksheetFoundation · PDFMark SchemeFoundation · PDF

Higher

WorksheetHigher · PDFMark SchemeHigher · PDF

Topic overview

When data is grouped into classes, the individual values are unknown, so the mean can only be estimated.

The method assumes every value in a class sits at the midpoint of that class. You multiply each midpoint by its frequency, add those products, then divide by the total frequency.

The answer is an estimate precisely because of that assumption. The true mean would need the original values, which grouping has discarded, so questions always ask for an estimate of the mean rather than the mean itself.

Revision notes

Finding the midpoints

Add the lower and upper bounds of each class and halve.

For the class \(10 \le x < 20\), the midpoint is \(15\). Getting the midpoints right is the foundation of the whole calculation.

Multiplying and totalling

Multiply each midpoint by its frequency, then add all the products.

This gives the estimated total of all the data values, which is what the mean calculation needs.

Dividing

Divide the total of the products by the total frequency.

If the products sum to \(1240\) and the total frequency is \(40\), the estimated mean is \(31\).

Key points

  • The mean from grouped data is an estimate.
  • Use the midpoint of each class.
  • Midpoint is the average of the class bounds.
  • Multiply each midpoint by its frequency.
  • Add the products, then divide by the total frequency.
  • It is an estimate because exact values are unknown.

Worked examples

Example 1

Find the midpoint of the class \(20 \le x < 30\).

Working

\[\frac{20 + 30}{2}\]average the class bounds
\[= 25\]state the midpoint

Example 2

Midpoints \(5, 15, 25\) have frequencies \(4, 6, 10\). Find the estimated mean.

Working

\[5(4) + 15(6) + 25(10) = 360\]multiply each midpoint by its frequency and add
\[4 + 6 + 10 = 20\]find the total frequency
\[360 \div 20 = 18\]divide to find the estimated mean

Example 3

Explain why the mean from a grouped table is only an estimate.

Working

\[\text{Exact values are not known}\]grouping discards the individual data
\[\text{Midpoints are assumed}\]so the answer is approximate

Common mistakes

  • Using the class boundaries instead of midpoints.

    Each class is represented by its midpoint, not its upper or lower bound.

  • Dividing by the number of classes.

    Divide by the total frequency, not by how many classes there are.

  • Forgetting to multiply by the frequency.

    Each midpoint counts as many times as its frequency.

  • Calling the answer the mean rather than an estimate.

    It is an estimate, and questions expect that word.

Exam tips

  • Add a midpoint column and a midpoint times frequency column to the table.
  • Check the midpoints are halfway between the class bounds.
  • Divide by the total frequency, not the number of classes.
  • Describe your answer as an estimate of the mean.

Key terms

Grouped data
Data sorted into classes rather than individual values.
Midpoint
The value halfway through a class.
Class
One group in a grouped frequency table.
Estimate
An approximate value based on assumptions.

Related topics

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.