Averages and weighted data

The mean is a total in disguise

The arithmetic mean is the total divided by the count, so every average hides a total:

$$\text{total} = \text{mean} \times \text{count}$$

This one line solves most average questions in data insight. To add or remove a value, go to the total, change it, and divide by the new count. If eight sales figures average 50, their total is 400; removing a figure of 22 leaves 378 over seven figures, a mean of 54.

Combining groups: weight by size

A table often gives an average for each group and the size of each group. The overall average is not the average of the averages unless the groups are the same size:

$$\text{overall mean} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2 + \dots}{n_1 + n_2 + \dots}$$

Twenty students averaging 70 and thirty averaging 80 have a combined total of $1\,400 + 2\,400 = 3\,800$ over 50 students: a mean of 76, not 75. The combined mean always lies between the group means, closer to the larger group. That fact alone often rules out options without any arithmetic.

A quick way to compute a weighted mean by hand is to measure from the lowest group. With weights of 30%, 10% and 60% on averages of 2,400, 3,200 and 2,000, the mean is $2\,000 + 0.3 \times 400 + 0.1 \times 1\,200 = 2\,000 + 120 + 120 = 2\,240$.

Rates combine the same way

Pass rates, conversion rates and prices per unit are averages too. A group's overall rate is the sum of the successes over the sum of the bases, never a simple average of the rates. A branch with 90% on 100 cases and another with 60% on 50 cases have $\frac{90 + 30}{150} = 80\%$ overall, not 75%.

When the overall comparison reverses

Weighting has a striking consequence. Two teams win contracts with small and large clients:

Team	Small clients	Large clients	Overall
P	9 of 10 (90%)	20 of 50 (40%)	29 of 60
Q	40 of 50 (80%)	3 of 10 (30%)	43 of 60

P does better with each kind of client, yet Q wins more contracts overall, because Q's cases are mostly the easy kind. So a statement such as "P has the higher success rate overall" cannot be judged from the subgroup rates alone; you need the sizes. When a display gives both, check both.

Means and medians from a frequency table

A frequency table lists each value and how many times it occurs. The mean is $\frac{\sum (\text{value} \times \text{frequency})}{\sum \text{frequency}}$. The median is the middle position in order: build running totals of the frequencies and see where the middle falls. The mode is the value with the highest frequency.

The mean and the median answer different questions. A few very large values pull the mean up but barely move the median, so with skewed data (incomes, house prices, waiting times) the mean is usually above the median. A question that says "typical" or "middle" wants the median; "per person" or "on average" with a total in view wants the mean.

What an average cannot tell you

An average alone says nothing about any single member of the group. A department averaging €2,400 may have nobody paid exactly €2,400, and a rise in the average does not mean everyone earned more: hiring one well-paid person can lift it. Statements of the form "every employee", "at least one" or "most employees" usually cannot be decided from averages, and in a data question that makes them the "cannot be determined" option, not the true one.

A routine for averages

Write each group as count, mean and total in three columns on your sheet. Add the counts and totals; divide once at the end. Before choosing, check that the result lies between the group means and leans towards the biggest group.