Grouped data estimates

When only class counts are available, the mean, SD and median are estimated by treating each class as concentrated at its midpoint and interpolating within the median class.

\xbar \approx \frac{\sum f_k m_k}{n}, \qquad s^2 \approx \frac{\sum f_k (m_k - \xbar)^2}{n-1}

Grouped mean and sample variance.

$f_k$: Count (frequency) in class $k$ .
$m_k$: Midpoint of class $k$ : (lower + upper)/2.
$n$: Total count $\sum f_k$ .

\text{median} \approx L + \frac{n/2 - F}{f_m}\,w

Grouped median by linear interpolation.

$L$: Lower boundary of the class containing the median.
$F$: Cumulative count before that class.
$f_m,\ w$: Count and width of the median class.

When to use

Published tables, frequency distributions or histograms where the raw values are not available.

Assumptions

Values are spread evenly within each class (for the median) or centred at the midpoint (for the mean and SD). Classes are contiguous and non-overlapping.

How to interpret

The results are estimates: the true mean and SD of the raw data may differ. The modal class is the class with the largest count (compare counts per unit width if widths differ).

Common mistakes

Using open-ended classes ("60+") without choosing a sensible upper limit — the midpoint is unknown.
Reporting grouped estimates with more decimals than the grouping justifies.
Using class limits instead of midpoints.

Example

Classes 0–10 (4 values), 10–20 (10), 20–30 (6); $n = 20$ . Midpoints 5, 15, 25: $\xbar \approx (20 + 150 + 150)/20 = 16$ ; $s \approx 7.18$ . Median: $n/2 = 10$ falls in 10–20 with $F = 4$ : $10 + (10 - 4)/10 \times 10 = 16$ . Modal class 10–20.