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Mathematics · Dispersion of grouped data

Estimating the mean from grouped data

The table hides the exact marks, so it is not clear which number to use for each class.

To estimate the mean from a grouped table, multiply each midpoint by its frequency, add the products, and divide by the total frequency. The answer is an estimate, because the exact values are not in the table.

This lesson is part of dispersion of grouped data. It uses the midpoints from reading class intervals and cumulative frequency.

What is the method?

  1. Find the midpoint x of each class.
  2. Multiply the midpoint by the frequency to get fx.
  3. Add the fx column to get Σfx.
  4. Divide Σfx by the total frequency Σf.

Worked example: marks of 40 students

Marks Midpoint x Frequency f fx
10 to 19 14.5 4 58
20 to 29 24.5 9 220.5
30 to 39 34.5 14 483
40 to 49 44.5 8 356
50 to 59 54.5 5 272.5
Total 40 1 390

Check each product: 4 × 14.5 = 58, 9 × 24.5 = 220.5, 14 × 34.5 = 483, 8 × 44.5 = 356 and 5 × 54.5 = 272.5. The sum is 58 + 220.5 + 483 + 356 + 272.5 = 1 390.

The mean is 1 390 ÷ 40 = 34.75.

Reasonableness check. The modal class is 30 to 39, with midpoint 34.5. The mean is close to it and lies between 14.5 and 54.5. The answer is sensible.

The slip that costs the most marks

The common slip is to add the midpoints and divide by the number of classes: (14.5 + 24.5 + 34.5 + 44.5 + 54.5) ÷ 5 = 34.5. This happens to be close here because the frequencies are roughly balanced, which hides the error.

When the frequencies are lopsided, the wrong method gives a clearly wrong answer. Try frequencies 1, 1, 1, 1 and 36. Nearly everyone is in the last class, yet the wrong method still gives 34.5. Always multiply by frequency.

Why the answer is only an estimate

Suppose the 14 students in 30 to 39 all scored 30. Then the true mean would be smaller than your estimate. If they all scored 39, it would be larger.

You cannot know which, so the paper asks for an estimate. Write the word “estimate” or “approximately” in your answer when the question uses it.

Check yourself

Waiting times at a clinic have frequencies 6, 11, 8 and 3 in the classes 1 to 5, 6 to 10, 11 to 15 and 16 to 20 minutes. Estimate the mean waiting time, to two decimal places.

Answer

Midpoints: 3, 8, 13, 18. Products: 6 × 3 = 18, 11 × 8 = 88, 8 × 13 = 104, 3 × 18 = 54. Σfx = 18 + 88 + 104 + 54 = 264. Σf = 28.

Mean = 264 ÷ 28 = 9.43 minutes (estimate). It sits inside the modal class 6 to 10, which is sensible.

What to study next

Continue with constructing and interpreting an ogive, which uses the cumulative column instead of the midpoints. Check your arithmetic on invented data using the descriptive statistics explorer.

For a teacher to watch your fx column, see online one-to-one Mathematics tuition.

Common questions

Why do we use the midpoint?

The exact values inside a class are not known. The midpoint is the best single value to stand for the whole class, because some values fall above it and some below. This is why the answer is an estimate and not the exact mean.

What is the formula for the mean of grouped data?

Mean = Σfx ÷ Σf, where x is the midpoint of each class and f is its frequency. Σfx is the sum of all the products, and Σf is the total frequency. Do not divide by the number of classes.

Can the estimated mean lie outside the modal class?

Yes. The mean uses every class, so a class far from the mode can pull it away. It should still lie between the lowest and highest midpoints. If it does not, an addition or a division has gone wrong.

Should I use class limits or boundaries for the midpoint?

Either gives the same midpoint when the classes are consecutive and have no gaps. Use the limits, because they are faster to add. Do not use the boundaries of one class and the limits of another.

If the method is clear but one product or total slips each time, one-to-one lessons let a teacher watch your fx column and find the line where the arithmetic or the midpoint goes wrong.

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