Statistical thinking for growth: why averages mislead you

One number standing in for thousands

Open almost any shop dashboard and the first row is a set of averages: order value, revenue per customer, conversion rate. Each squeezes thousands of orders into one figure, which is why it is useful. Nobody can hold 2,000 orders in their head.

But a summary works by throwing things away. The average keeps the centre of gravity of the data and discards the shape: how spread out it is, whether it leans to one side, whether it is really two different groups of customers sharing one column. In growth work the shape is usually where the decision sits. The average does not lie on purpose. It lies by leaving out.

Below are the handful of ideas that stop an average from misleading you. None needs more than school arithmetic.

Throughout, I use one invented business. Harbourside Ceramics is a small pottery shop in Cork, run by Ciarán, selling mugs, bowls and plates online, plus the odd bulk order from cafés around the county. Ciarán is nobody real, and every number in this article is made up so I can show every step.

The mean answers a question you probably did not ask

The arithmetic mean is what most people mean by "the average": add up all the values and divide by how many there are.¹

$$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$$

Here $\bar{x}$ (read "x bar") is the mean, $n$ is the number of values, $x_i$ is the $i$th value, and $\sum_{i=1}^{n}$ means "add up every value from the first, $i = 1$, to the last, $i = n$". The idea is built up from scratch in Foundations: the mean.

The question the mean answers is precise: if the total were shared out equally, how much would each one get? That is a perfectly good question for some jobs. It is rarely the question "what does a normal customer spend?", which is the one people think they are asking.

Worked example: ten orders on a Tuesday

Ciarán takes ten orders on a quiet Tuesday. Sorted from smallest to largest, in euro:

Order	1	2	3	4	5	6	7	8	9	10
Value	€22	€24	€24	€28	€30	€32	€35	€38	€45	€420

The last one is a café in Douglas ordering 40 mugs for its new counter.

Step one, add them up: $22 + 24 + 24 + 28 + 30 + 32 + 35 + 38 + 45 + 420 = 698$.

Step two, divide by the number of orders, $n = 10$: $\bar{x} = 698 / 10 = 69.8$.

The mean order on Tuesday was €69.80. Now look at the table again. Nine of the ten orders are €45 or less. Not one customer spent anything close to €69.80. The figure is pulled up by a single order, and it describes nobody.

The median and the mode describe the typical order

Two other summaries answer the question people actually mean. The median is the middle value once everything is sorted: half the orders are below it and half above. With an even number of values, it is halfway between the two in the middle. The mode is the value that turns up most often.¹

The median is the better measure of the centre when there are extreme values, because it only cares about the order of the values, not their size. Make the biggest order twice as big and the median does not move.¹ Both are explained in Foundations: the median and the mode.

Worked example: the median and mode of Ciarán's Tuesday

The ten orders are already sorted. With $n = 10$, the middle sits between the fifth and the sixth value, which are €30 and €32, so the median is $(30 + 32) / 2 = 31$, or €31.

The mode is €24, the only value that appears twice. It is the price of Ciarán's bestselling pair of mugs.

So Tuesday reads: mean €69.80, median €31, mode €24. The median is the one that matches what a typical customer did.

Now take the café order out and do it again. Nine orders, total $698 - 420 = 278$, so the mean becomes $278 / 9 = 30.89$. The median of nine sorted values is the fifth one, €30. One order moved the mean from €30.89 to €69.80, more than doubling it. The same order moved the median by €1.

When the mean sits above the median, the data leans right

The order mean, then median, then mode, from largest to smallest, is not a coincidence. It is the signature of a distribution that is skewed to the right: most values bunched at the low end and a long tail stretching out towards the big ones. In a distribution skewed to the right, the mean is usually greater than the median, and the mode is the smallest of the three.²

A distribution is the full picture of how often each value turns up, what you see when you sort orders into bands and count each band. It is covered in Foundations: distributions.

In the shops I have worked on, order values lean to the right almost every time: lots of small and middling baskets, a thinner stream of large ones, and now and then a trade order several times the size of anything else. The mean of data shaped like that is not wrong, but it sits out in the tail, away from most customers. The official reference handbook of the US National Institute of Standards and Technology (NIST) puts it plainly: for skewed data the mean and the median do not give similar estimates of location, and extreme values in the tails distort the mean.³

Worked example: a whole quarter of orders

Ciarán's last quarter brought 2,000 orders and €132,000 of revenue. Sorted into value bands, they look like this:

Across the bottom are the value bands in euro, from under €20 on the left to over €500 on the right. Up the side is the number of orders in each band. The peak is the €20 to €30 band, then the bars fall away step by step into a long, thin tail. That is the shape of a right skew. The takeaway: the bulk of Ciarán's customers live in the first four bars, and the mean lives somewhere else.

Now the three summaries, step by step.

The mean is the total divided by the count: $132{,}000 / 2{,}000 = 66$, so €66.

The median is the value of the middle order, between the 1,000th and the 1,001st in sorted order. Count up the bars: 180 orders under €20, then 520 more takes us to 700, then 420 more takes us to 1,120. So the middle orders sit inside the €30 to €40 band. Ciarán looks them up in his order export: the median is €36.

The mode is €24, his bestselling pair of mugs again.

Mean €66, median €36, mode €24. Counting the bars once more, everything under €60 is $180 + 520 + 420 + 380 = 1{,}500$ orders, and a few dozen more sit between €60 and €66. So roughly 1,540 of the 2,000 orders, about 77%, come in below the mean. "Our average order is €66" is true. "Our customers spend about €66" is not.

Worked example: the free delivery threshold

Here is where it costs money. Ciarán wants to offer free delivery above a threshold, to nudge people into adding one more item. His first instinct is to set it just above the average: €70.

The useful question is not "where is the average?" but "how many orders sit a short reach below the line?". A customer €8 short might add a mug. A customer €40 short will not add three. Ciarán picks €15 as a reasonable reach.

For a €70 threshold, the orders a reach below are those between €55 and €70. From his export that is about 160 orders, 8% of the quarter. The typical customer, at €36, is €34 away and simply pays for delivery.

For a €45 threshold, the reach covers €30 to €45: about 515 orders, roughly a quarter of all orders.

Which line is right depends on his margin and delivery costs, because a lower line also gives free delivery to everyone already above it. But the mean cannot even start that calculation. Only the distribution tells Ciarán who is near enough to reach.

Spread tells you whether to trust the centre

A centre without a spread is half a sentence. Two shops can both have a mean order of €66, one with nearly every order between €55 and €75, one with orders from €8 to €2,000. Only the first can plan off the mean.

The standard deviation measures how far the values typically are from their mean.⁴ For a sample it is calculated like this:

$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} \left(x_i - \bar{x}\right)^2}$$

Here $s$ is the sample standard deviation, $n$ is the number of values, $x_i$ is the $i$th value, $\bar{x}$ is their mean, and $\left(x_i - \bar{x}\right)^2$ is the squared distance of each value from the mean. Squaring makes every distance positive and weights big distances heavily. Dividing by $n - 1$ rather than $n$ gives a better estimate of the spread in the whole population when you only have a sample of it.⁴ The square root brings the result back into euro. Without the square root the quantity is called the variance, $s^2$. All of it, including why it is $n - 1$, is in Foundations: spread.

A handy companion is the coefficient of variation, the standard deviation divided by the mean:

$$CV = \frac{s}{\bar{x}}$$

Here $CV$ is the coefficient of variation, $s$ is the standard deviation and $\bar{x}$ is the mean. Because the euro cancels out, it lets you compare the spread of things measured on different scales. The same Foundations section on spread explains it, along with a rule of thumb I use: once the $CV$ is above 1, the mean on its own is not a safe summary, and I always report the median next to it.

Worked example: the spread of Ciarán's Tuesday

Back to the ten orders, with $\bar{x} = 69.8$.

Step one, the distance of each order from the mean, $x_i - \bar{x}$: the nine small orders come out between minus €47.80 (for €22) and minus €24.80 (for €45). The café order comes out at $420 - 69.8 = 350.2$.

Step two, square each distance. The nine small ones give 2,284.84, 2,097.64, 2,097.64, 1,747.24, 1,584.04, 1,428.84, 1,211.04, 1,011.24 and 615.04. The café gives $350.2^2 = 122{,}640.04$.

Step three, add the squares: $136{,}717.6$.

Step four, divide by $n - 1 = 9$: $136{,}717.6 / 9 = 15{,}190.84$. That is the variance, $s^2$.

Step five, take the square root:

$$s = \sqrt{15{,}190.84} = 123.25$$

So $CV = 123.25 / 69.8 = 1.77$. Well above 1: report the median.

Look at step three again. The café order alone contributes 122,640.04 of the 136,717.6, almost 90%. One order is the spread. Without it, the nine orders have a mean of €30.89, a standard deviation of €7.51 and a $CV$ of 0.24: a grand, steady little shop with one café on top.

That is the catch with the standard deviation: it is built on the mean, so it inherits the mean's weakness. Outliers can make it very large, and in skewed data it may not be much help.⁴ NIST notes the same: extreme values in the tails distort it.⁵

Quartiles: a spread that outliers cannot push around

The answer for skewed data is to describe the spread with positions rather than distances. Percentiles divide sorted data into 100 equal parts, and quartiles divide it into four. The first quartile, $Q_1$, has a quarter of the values below it. The second is the median. The third, $Q_3$, has three quarters below it. The distance between the first and third quartile is the interquartile range, the spread of the middle half of the data.⁶

$$IQR = Q_3 - Q_1$$

Here $IQR$ is the interquartile range, $Q_3$ is the third quartile and $Q_1$ is the first quartile. Because it only looks at the middle half, the biggest and smallest values cannot touch it. NIST lists it among the measures of spread that hold up when the tails are extreme.⁵ The median it rests on is covered in Foundations: the median and the mode.

The same quartiles give a standard test for values worth a second look. A value is a potential outlier if it sits more than $1.5 \times IQR$ below $Q_1$ or more than $1.5 \times IQR$ above $Q_3$.⁶

Worked example: the quartiles of Tuesday

Split the ten sorted orders at the median into a lower five (€22, €24, €24, €28, €30) and an upper five (€32, €35, €38, €45, €420). The first quartile is the median of the lower five, €24. The third quartile is the median of the upper five, €38. (Spreadsheets use slightly different quartile rules, so yours may differ by a euro or two.)

$$IQR = 38 - 24 = 14$$

The middle half of Tuesday's orders spans just €14. The upper fence is $38 + 1.5 \times 14 = 38 + 21 = 59$. The café order, at €420, is far above it, so it is flagged. The lower fence is $24 - 21 = 3$, and nothing is below that.

Worked example: Ciarán's quarter in one line

For the full quarter, Ciarán counts through the sorted export the same way. The 500th order (the first quartile) is about €26. The 1,500th (the third quartile) is €60. The 1,800th, the 90th percentile, is about €110.

His quarterly summary now reads: "2,000 orders, €132,000 revenue. Typical order €36, middle half between €26 and €60, one order in ten above €110. Mean €66." Every figure on that line tells him something the mean alone could not.

Outliers are usually customers, not mistakes

An outlier test flags a value. It does not tell you what to do with it. There are three honest possibilities, and only one of them is "delete it".

It can be an error: a test order, a duplicated import, a refund booked as a sale. Fix it at the source and remove it.

It can be a genuine but rare event: a one off order from a wedding planner. Keep it, but report a centre that it cannot drag around. The median does this. So does a trimmed mean, which drops a fixed share of values from each end before averaging. NIST gives 5% at each end as a common choice and groups the trimmed mean with other summaries that are not unduly affected by the tails.³

Or it can be a different kind of customer altogether. That is the most valuable case, and the one most often missed. The café is not a strange retail customer. It is a trade customer, buying for a business, at a different scale, for different reasons. Mixing the two into one average describes neither.

Worked example: a trimmed mean, then a proper split

On Tuesday's ten orders, trimming one order from each end (10% each side, the gentlest trim possible with ten values) drops €22 and €420. The remaining eight total $698 - 22 - 420 = 256$, so the trimmed mean is $256 / 8 = 32$. Close to the median of €31, as you would expect.

Then the split. Ciarán goes through the quarter and tags his trade orders. There are 60 of them, all above €200, worth €33,000 together, so the mean trade order is $33{,}000 / 60 = 550$, or €550.

Retail is everything else: $2{,}000 - 60 = 1{,}940$ orders worth $132{,}000 - 33{,}000 = 99{,}000$. The mean retail order is $99{,}000 / 1{,}940 = 51.03$, about €51. The retail median barely moves, at about €35.

Ciarán does not have one business with a mean order of €66. He has a retail shop with a typical order of €35 and a small trade line around €550, which need different prices, emails and delivery offers. The single average blended them into a customer who does not exist.

The mean describes no segment at all

The same blending happens with customers over a year, and it is where the money hides. When customers fall into groups that behave very differently, the overall mean is a weighted mixture of the group means:

$$\bar{v} = \frac{\sum_{s=1}^{S} n_s \, v_s}{\sum_{s=1}^{S} n_s}$$

Here $\bar{v}$ is the overall mean value per customer, $S$ is the number of segments, $s$ counts through them, $n_s$ is the number of customers in segment $s$, and $v_s$ is the mean value per customer in that segment. The top line is the total value, the bottom line is the total number of customers. It is the ordinary mean from Foundations: the mean, just added up group by group.

Worked example: Ciarán's customers over a year

Ciarán sorts last year's 1,500 retail customers by what each spent in the year and cuts them into three groups.

Segment	Customers	Share of customers	Spend per customer	Revenue	Share of revenue
One order and gone	1,000	67%	€30	€30,000	29%
Came back once	350	23%	€90	€31,500	30%
Regulars	150	10%	€280	€42,000	41%
All	1,500	100%	€69	€103,500	100%

The revenue column is customers times spend: $1{,}000 \times 30 = 30{,}000$, $350 \times 90 = 31{,}500$ and $150 \times 280 = 42{,}000$. The total is $103{,}500$, and the mean per customer is $103{,}500 / 1{,}500 = 69$.

€69 per customer. No group spends anything like it. Two thirds of the customers spend €30, and one in ten spends €280. The regulars are 10% of the people and 41% of the revenue.

A tempting conclusion is that a few big customers beat many small ones. Careful: ten customers at €100 and a hundred at €10 bring exactly the same €1,000. What differs is what each group costs to win and to serve, and what lifts it. The regulars respond to early access and a reason to come back. The one order customers need a reason to place a second order at all. Treating them all as the €69 customer gets neither right. How to find groups like these properly, and which methods to use, is the subject of segmentation strategy before algorithms and the tools of segmentation.

Simpson's paradox: every part improves and the whole gets worse

The strangest way an average misleads is when it points the wrong way entirely. Simpson's paradox is the name for an association that appears, disappears or reverses when a population is split into groups.⁷ It happens whenever the mix of groups changes between the things you compare.

The classic case is graduate admissions at the University of California, Berkeley, studied by Bickel and colleagues in 1975. Across the university, men were admitted at a higher rate than women. Department by department, there was no bias against women. The difference came from where people applied: women applied more often to the departments that were hardest to get into.⁷ The overall rate was a mixture, and the mixture differed.

In a shop the same thing hides inside conversion rates, which are ratios of orders to visits (see Foundations: ratios and rates). The overall rate is a weighted mean of the segment rates, weighted by each segment's share of the traffic:

$$p = \sum_{s=1}^{S} w_s \, p_s$$

Here $p$ is the overall conversion rate, $S$ is the number of segments, $w_s$ is segment $s$'s share of all visits (the shares add up to 1), and $p_s$ is the conversion rate inside segment $s$. It is the weighted form of the mean in Foundations: the mean.

Worked example: November looks worse, and is better

Ciarán runs a social media campaign in November that brings in a wave of visitors on phones. He compares conversion with October, split into desktop and mobile:

Month	Device	Visits	Orders	Conversion rate
October	Desktop	10,000	400	4.0%
October	Mobile	10,000	150	1.5%
October	All	20,000	550	2.75%
November	Desktop	6,000	252	4.2%
November	Mobile	24,000	384	1.6%
November	All	30,000	636	2.12%

Desktop went up, from 4.0% to 4.2%. Mobile went up, from 1.5% to 1.6%. The overall rate went down, from 2.75% to 2.12%.

The formula shows why. In October each device had half the visits, so $w_{\text{desktop}} = w_{\text{mobile}} = 0.5$, and $p_{\text{Oct}} = 0.5 \times 4.0\% + 0.5 \times 1.5\% = 2.0\% + 0.75\% = 2.75\%$.

In November desktop had $6{,}000 / 30{,}000 = 0.2$ of the visits and mobile $24{,}000 / 30{,}000 = 0.8$, so $p_{\text{Nov}} = 0.2 \times 4.2\% + 0.8 \times 1.6\% = 0.84\% + 1.28\% = 2.12\%$.

The weights moved towards the device that always converts less. Both rates inside the segments improved, and the mix shift swamped them.

Across the bottom are the two months. Up the side is the conversion rate, as a percentage. The desktop and mobile lines both slope gently upwards, and the line for all visitors slopes down between them. The takeaway: an overall rate can fall while every part of the business improves, simply because the traffic mix moved.

There is a second trap in the same table. If Ciarán averages the two November rates directly, $(4.2\% + 1.6\%) / 2 = 2.9\%$, he gets a number that is neither the true overall rate of 2.12% nor anything else real. An average of averages ignores the weights. Always go back to the totals: 636 orders divided by 30,000 visits.

The split view is not always the right one. If the campaign brought in visitors who will never buy, the lower total is the honest answer to "was it worth it?". The paradox does not say which view to trust. It says look at both, and ask why the mix changed. For comparing two rates properly, with the uncertainty included, see statistical testing, sample size and lift.

When the mean is exactly the right number

A word in the mean's defence: it answers the "shared out equally" question, and some decisions are exactly that question.

Totals are one. Mean times count is the total, the only centre with that property. If Ciarán expects 2,200 orders next quarter and wants a revenue forecast, 2,200 orders at €66 each, €145,200, is right, and 2,200 at the median of €36 would be badly wrong. Budgets, cash flow and stock in euro all want the mean.

Expected value is another. When you weigh options by their likely payoff, you multiply outcomes by probabilities and add them up, which is a mean (see Foundations: expected value and decision theory for founders).

And comparing revenue per visitor in a test is a third. Individual orders are skewed, but the mean of many orders behaves far more predictably than any single order, which is the central limit theorem described in Foundations: distributions. With enough orders, comparing means is sound. With thirty orders and one café, it is not.

Worked example: which number for which decision

Ciarán has three decisions this month. For the revenue forecast for the bank, he uses the mean, €66, times the expected orders. For the free delivery threshold, he uses the distribution and the count of orders a short reach below each line. For the retail email, he uses the median retail order, about €35, and the products around it, because that is what a typical reader buys. Three decisions, three numbers, one set of orders.

A routine I use before trusting any average

None of this needs special software, just a habit. Before an average goes into a report or a decision, I run this short routine.

The diagram reads from the top: the decision first, then the shape, then skew, outliers and hidden groups, and only then the number.

Worked example: Ciarán's monthly report, rewritten

Before, Ciarán's monthly report had one line on orders: "Average order value €66, down 3% on last month."

After the routine, it has three short lines:

"Retail: 616 orders, typical order €35 (middle half €26 to €58), mean €51. Trade: 20 orders, mean €550. Conversion by device: desktop 4.2%, mobile 1.6%, overall 2.12% because mobile traffic grew."

It takes five minutes longer. He no longer loses a morning giving out about a 3% drop that was really three fewer café orders.

The point in one sentence

An average is a summary, and every summary throws something away, so before you act on one, look at the shape it came from, put the median and a spread next to it, and check whether the total hides groups that should be compared on their own.

If you want the arithmetic behind the ideas here without the shop around it, start with Foundations, and for the wider toolkit, the statistics every marketer actually needs.

Footnotes

1. Barbara Illowsky and Susan Dean, Introductory Statistics 2e, section 2.5 Measures of the Center of the Data, OpenStax, Rice University, Houston, 2023, https://openstax.org/books/introductory-statistics-2e/pages/2-5-measures-of-the-center-of-the-data

2. Barbara Illowsky and Susan Dean, Introductory Statistics 2e, section 2.6 Skewness and the Mean, Median, and Mode, OpenStax, Rice University, Houston, 2023, https://openstax.org/books/introductory-statistics-2e/pages/2-6-skewness-and-the-mean-median-and-mode

3. NIST/SEMATECH, e-Handbook of Statistical Methods, section 1.3.5.1 Measures of Location, National Institute of Standards and Technology, https://www.itl.nist.gov/div898/handbook/eda/section3/eda351.htm

4. Barbara Illowsky and Susan Dean, Introductory Statistics 2e, section 2.7 Measures of the Spread of the Data, OpenStax, Rice University, Houston, 2023, https://openstax.org/books/introductory-statistics-2e/pages/2-7-measures-of-the-spread-of-the-data

5. NIST/SEMATECH, e-Handbook of Statistical Methods, section 1.3.5.6 Measures of Scale, National Institute of Standards and Technology, https://www.itl.nist.gov/div898/handbook/eda/section3/eda356.htm

6. Barbara Illowsky and Susan Dean, Introductory Statistics 2e, section 2.3 Measures of the Location of the Data, OpenStax, Rice University, Houston, 2023, https://openstax.org/books/introductory-statistics-2e/pages/2-3-measures-of-the-location-of-the-data

7. Jan Sprenger and Naftali Weinberger, Simpson's Paradox, Stanford Encyclopedia of Philosophy, 2021, revised 2026, https://plato.stanford.edu/entries/paradox-simpson/