Grouped data and the geometric mean

Midpoints for binned data, the right average for growth rates, and which centre and spread to report for skewed data.

When only the table survives

Sometimes the raw values are gone and all you have is a frequency table. Treat every value in a bin as sitting at the bin's midpoint $m$ , weight by the frequency $f$ , and the mean and SD formulas become

$\overset{x}{ˉ} = \frac{\sum f m}{n}, s^{2} = \frac{\sum f m ^{2} - n x ˉ ^{2}}{n - 1}, n = \sum f$

The answers are approximations — the true values are not all at the midpoints — but they are the best the table allows, and the paper accepts them.

Worked example — days to pay, from a table

Northfield Credit published only a binned table of days to pay for 30 invoices.

Days	midpoint $m$	$f$	$f m$	$f m^{2}$
0 to under 10	5	6	30	150
10 to under 20	15	10	150	2,250
20 to under 30	25	8	200	5,000
30 to under 40	35	4	140	4,900
40 to under 50	45	2	90	4,050
		30	610	16,350

$\overset{x}{ˉ} = 610/30 = 20.33$ days. $s^{2} = \frac{16 , 350 - 30 \times 20.3 3 ^{2}}{29} = \frac{16 , 350 - 12 , 403.3}{29} = 136.09$ , so $s = 11.67$ days. Notice the divisor: $n - 1 = 29$ , exactly as for raw data.

The right average for a growth rate

A fund returned +12 %, −5 %, +8 % and +20 % over four years. The arithmetic mean, $(0.12 - 0.05 + 0.08 + 0.20) /4 = 8.75 %$ , is not the rate that would have produced the same final value. Growth compounds — multiplies — so the average that matters is the geometric mean of the growth factors:

$\overset{g}{ˉ} = (1.12 \times 0.95 \times 1.08 \times 1.20)^{1/4} - 1 = (1.37894)^{1/4} - 1 = 0.0836$

An 8.36 % rate every year for four years gives the same $1.379 multiple that the actual sequence did. The arithmetic mean always overstates it (equal only when every year's return is identical), because a loss and a gain of the same size do not cancel under multiplication: −5 % then +5 % leaves you at 99.75 %.

Which centre, which spread

The module has given you two pairs. Mean and SD — efficient, the basis of every inference formula, and sensitive to outliers. Median and IQR — resistant, and the honest description of a skewed distribution with a long tail or a few extreme values. For Maple Ledger's overdue days (right-skewed, with a 41), report median 8 and IQR 11; the mean 11.27 is being dragged by one account. For the near-symmetric audit hours, mean 44 and SD 5.29 are fine. The paper marks the choice as well as the arithmetic.