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A short class with Sarah and her students. The other ways are under More.
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We know 2 students scored 0–10, 3 scored 10–20 and 5 scored 20–30, but not their exact marks. What value should stand for each class?
The midpoint: 5, 15 and 25.
Now multiply each midpoint by its frequency and add.
10 + 45 + 125 = 180. There are 10 students, so the mean is 18.
Is 18 the exact mean?
No, it's an estimate, because we pretended everyone in a class scored the midpoint.
Find each midpoint's deviation from 18 and square it.
−13, −3 and 7, so the squares are 169, 9 and 49.
Each square counts once for every student in that class.
So 2 × 169 = 338, 3 × 9 = 27 and 5 × 49 = 245. They add up to 610.
Divide by the total frequency and take the square root.
610 ÷ 10 = 61, and √61 is about 7.81 marks.
Let's check with σ² = Σfx² ÷ Σf − x̄². What is Σfx²?
2 × 25 + 3 × 225 + 5 × 625, which is 3850.
3850 ÷ 10 is 385, minus 18 squared, 324, gives 61. It matches!
Now suppose 5 students scored 0–10 and 5 scored 20–30, with none in the middle. Which group is more spread out?
The new one. All its values are far from the mean of 15, so σ comes out as 10.
each class is replaced by its midpoint, weighted by its frequency.
x̄ = Σfx ÷ Σf; here 180 ÷ 10 = 18.
σ = √(Σf(x − x̄)² ÷ Σf); here √61 ≈ 7.81.
σ² = Σfx² ÷ Σf − x̄² gives the same variance with less subtraction.
grouped results are approximate because the exact values within each class are unknown.
Marks 0–10 (2), 10–20 (3), 20–30 (5): midpoints 5, 15, 25, Σfx = 180, x̄ = 18, Σf(x − x̄)² = 610, σ = √61 ≈ 7.81.
50 students in classes 0–20 (5), 20–40 (10), 40–60 (15), 60–80 (12), 80–100 (8): x̄ = 2660 ÷ 50 = 53.2 and σ = √(29 088 ÷ 50) ≈ 24.1.
Waiting times 0–5 (4), 5–15 (10), 15–30 (6): midpoints 2.5, 10, 22.5, x̄ = 245 ÷ 20 = 12.25 and σ = √(4062.5 ÷ 20 − 12.25²) ≈ 7.28 minutes.
Marks 0–10 (5) and 20–30 (5) only: x̄ = 15 and every midpoint is 10 from the mean, so σ = 10, more spread than the first example.
Middle of each class, times how many there are; add and divide for the mean, then square each gap from afar.
A teacher lost the exact marks but kept the tally chart. She pretended every student in a class scored the middle mark, weighted each class by its tally, and recovered a mean of 18 and a spread of about 7.8.
"Mid, Multiply, Mean, Measure the gaps" - midpoints, multiply by f, find the mean, then the weighted squared deviations.
M.F.D.S.R. - Midpoints, Frequencies, Deviations, Squares (times f), Root.
Term
What value represents a class in grouped data?
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Term
What is the estimated mean of 0–10 (2), 10–20 (3), 20–30 (5)?
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Term
State the grouped-data standard deviation formula.
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What is the estimated standard deviation of 0–10 (2), 10–20 (3), 20–30 (5)?
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Why are grouped mean and standard deviation only estimates?
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Data organised into classes, with a frequency for each class instead of individual values.
A range of values forming one class, such as 10–20.
The middle of a class, (lower boundary + upper boundary) ÷ 2, used to represent the class.
The number of data values in a class.
x̄ = Σfx ÷ Σf, calculated from midpoints and frequencies.
σ = √(Σf(x − x̄)² ÷ Σf), an estimate of the spread of grouped data.
Σf, the total number of data values.
Example 2 estimates the mean and standard deviation of grouped marks (0–10: 2 students, 10–20: 3, 20–30: 5) using class midpoints. The mean is 180 ÷ 10 = 18 and the standard deviation is √(610 ÷ 10) = √61 ≈ 7.81 marks.
When data come in classes rather than as individual values, each class is represented by its midpoint and weighted by its frequency. For the marks 0–10 (frequency 2), 10–20 (3) and 20–30 (5), the midpoints are 5, 15 and 25, so the estimated mean is Σfx ÷ Σf = 180 ÷ 10 = 18. The squared deviations of the midpoints from 18, multiplied by the frequencies, add up to 610, so σ = √(610 ÷ 10) = √61 ≈ 7.81. This lesson sets out the full table, checks the answer with the Σfx² shortcut, explains why grouped results are only estimates, shows the calculator method, and works through further examples, including classes of unequal width and a comparison of two distributions.
In Example 1 every individual mark was known. In grouped data we only know how many values fall in each class, so we use each class's midpoint to stand for all the values in it.
The data. Marks of 10 students, grouped (for example, the class 0–10 means 0 ≤ mark < 10):
Step 1: midpoints. Midpoint = (lower boundary + upper boundary) ÷ 2, giving x = 5, 15 and 25.
Step 2: the estimated mean. Multiply each midpoint by its frequency: 2 × 5 = 10, 3 × 15 = 45 and 5 × 25 = 125. Then Σfx = 180, Σf = 10 and x̄ = 180 ÷ 10 = 18.
Step 3: deviations, squares and frequencies.
The last column adds up to 610.
Step 4: variance and standard deviation. Using the grouped-data formula σ = √(Σf(x − x̄)² ÷ Σf), σ² = 610 ÷ 10 = 61 and σ = √61 ≈ 7.81 marks.
A check with the shortcut formula. σ² = Σfx² ÷ Σf − x̄². Here Σfx² = 2 × 25 + 3 × 225 + 5 × 625 = 3850, so σ² = 3850 ÷ 10 − 18² = 385 − 324 = 61, the same answer.
Why the answers are estimates. We do not know the actual marks inside each class. Using midpoints assumes the values in each class are spread evenly around the middle, so the mean and standard deviation of grouped data are estimates of the true values.
Sample or whole group? Like the parent lesson, this calculation divides by Σf, treating the 10 students as the whole group. If they were a sample from a larger population, you would divide by Σf − 1 instead: s² = 610 ÷ 9 ≈ 67.8 and s ≈ 8.23.
Using a calculator. Enter the midpoints in one list and the frequencies in another, then run one-variable statistics with the frequency list. The calculator gives x̄ = 18, σx ≈ 7.81 and sx ≈ 8.23.
More examples.