How this instrument works
Sometimes raw data isn't available, only a summary table: a set of class intervals (often already reduced to their midpoints) and how many observations fell into each one — the kind of table a census report, a grouped survey result, or a grouped-frequency exam-score table typically publishes. You can still estimate the mean and standard deviation from a table like that, treating every observation inside a class as if it sat exactly at that class's midpoint. It's an approximation, since real values inside a class are spread across the whole interval rather than sitting precisely at its center, but it's the standard, widely taught method when only grouped data is available.
The mean is a frequency-weighted average of the midpoints: multiply each midpoint by its frequency, add those products together, and divide by the total frequency. The variance follows the same logic as an ordinary variance calculation, but every squared deviation from the mean is weighted by how many observations that class contributes: ss = Σ f·(midpoint − mean)². Dividing that sum by (N − 1) gives the sample standard deviation after taking a square root; dividing by N instead gives the population standard deviation.
This calculator needs two matching series — Class midpoints and Class frequencies — entered in the same order, one frequency per midpoint. It does not take raw, ungrouped data directly; for that, this site's plain standard-deviation instruments (which work from a full list of individual values rather than a midpoint/frequency table) are the right tool instead.
- Enter each class's representative value into Class midpoints, in order.
- Enter the matching count for each class into Class frequencies, in the same order as the midpoints.
- Read Grouped mean beneath — the frequency-weighted average of your midpoints.
- Read Sample standard deviation if your data is a sample standing in for a larger population (divides by N − 1).
- Read Population standard deviation if your grouped data already represents the entire population of interest (divides by N).
Worked example — an age-bracket survey of 45 people
A survey groups 45 respondents into five age brackets, represented by midpoints 5, 15, 25, 35, and 45, with 3, 7, 12, 15, and 8 people respectively in each bracket. Enter 5, 15, 25, 35, 45 into Class midpoints and 3, 7, 12, 15, 8 into Class frequencies. Total frequency N = 3+7+12+15+8 = 45; the weighted sum Σ(f·x) = 15+105+300+525+360 = 1,305, so Grouped mean = 1,305/45 = 29.0 exactly.
The weighted sum of squared deviations is ss = 3(24)²+7(14)²+12(4)²+15(6)²+8(16)² = 1,728+1,372+192+540+2,048 = 5,880. Sample standard deviation = √(5,880/44) ≈ 11.560, treating these 45 people as a sample; Population standard deviation = √(5,880/45) ≈ 11.431, treating them as the complete group of interest — the two figures differ only in whether the denominator is N−1 or N.
Questions
Why use midpoints instead of the actual raw values?
Because with grouped data, the actual raw values inside each class usually aren't available at all — only the class boundaries and a count of how many observations fell inside each one. The midpoint is the single best stand-in for every value in that class when the individual values are unknown, on the reasonable assumption that observations within a class are roughly evenly spread across it. If you do have the raw individual values, this site's plain standard-deviation instrument will give a more accurate result than the grouped-data approximation.
How accurate is the grouped-data estimate compared to the true standard deviation?
It's an approximation, and its accuracy depends on how narrow the classes are and how evenly the real values are actually spread within each one. Narrower classes (more of them, covering a smaller range each) generally produce a closer approximation to the true raw-data standard deviation, since the midpoint assumption introduces less error the tighter each class is. With only a handful of wide classes, the estimate can drift noticeably from the true figure.
Should I use sample or population standard deviation for grouped data?
The same rule applies as with ungrouped data: use sample standard deviation (dividing by N−1) when your grouped counts represent a sample standing in for a larger population you haven't fully measured, and population standard deviation (dividing by N) when the grouped counts already represent every member of the group you care about — a full class's exam scores, say, rather than a sampled subset of students.
What if my class frequencies list and midpoints list are different lengths?
The calculator requires one frequency value for every midpoint, entered in matching order — the first frequency pairs with the first midpoint, the second with the second, and so on. Mismatched list lengths trigger a validation message asking you to enter the same number of midpoints and frequencies before it will compute anything.
Does this work the same as plain standard deviation on a small, evenly spread data set?
Yes — as a check, entering midpoints 1, 2, 3, 4 with a frequency of 1 for each (equivalent to the plain data set 1, 2, 3, 4) gives a sample standard deviation of about 1.291, exactly matching what this site's ordinary standard-deviation instrument returns for that same raw data set. Grouped-data standard deviation is a generalization of the plain formula, not a different calculation.