NovaMarket's satisfaction survey is back from the field: 640 responses are sitting in your inbox, alongside last quarter's receipts and the e-commerce delivery logs. Marta stops by your desk and sums up the assignment in a single sentence: "Don't bring me 640 numbers; bring me one that represents them well." That is exactly what measures of central tendency are for: condensing a data set into one "typical" or "central" value. In this lesson you will learn the three main ones — mean, median and mode — their useful variants (the weighted, trimmed and geometric means) and, above all, the criteria for deciding which one to report to Marta on each occasion, because choosing badly is not a minor technical slip: it can change a business decision.

1. What is a measure of central tendency?
2. The arithmetic mean
3. The weighted mean
4. The median
5. The mode
6. The effect of extreme values and the trimmed mean
7. Which one do I use? Criteria for choosing
8. The geometric mean for growth rates

A measure of central tendency is a single number that tries to represent the "center" of a data set — the value around which the observations cluster. It is the most basic, and most widely used, form of statistical summary.

The fact that there are three main measures (rather than one) already tells you something important: the "center" of a data set is not a single concept. Each measure defines the center in a different way:

In many data sets the three roughly coincide and it hardly matters which one you use. But in NovaMarket's real data — with the occasional huge receipt, and satisfaction scores bunched up at the high end — they can drift far apart, and then the choice matters.

The arithmetic mean (or simply the mean) is the sum of all the values divided by the number of values. For a sample of \( n \) observations \( x_1, x_2, \dots, x_n \):

\[ \bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n} = \frac{\sum_{i=1}^{n} x_i}{n} \]

Recall the notation from the Basic Concepts lesson: \( \bar{x} \) (read "x-bar") is the mean of a sample (a statistic), while \( \mu \) is the mean of the population (a parameter).

Marta wants a first look at receipt amounts. You take a small sample of 10 receipts from the Madrid-Centro store from yesterday afternoon (in euros):

\[ 12.50 \quad 18.90 \quad 22.40 \quad 27.30 \quad 29.80 \quad 31.60 \quad 38.20 \quad 45.70 \quad 52.10 \quad 289.00 \]

(The €289.00 receipt is genuine: a local bar did its weekly stock-up.)

Step 1 — Add up all the values:

\[ 12.50 + 18.90 + 22.40 + 27.30 + 29.80 + 31.60 + 38.20 + 45.70 + 52.10 + 289.00 = \text{€}567.50 \]

Step 2 — Divide by the number of observations (\( n = 10 \)):

\[ \bar{x} = \frac{567.50}{10} = \text{€}56.75 \]

The mean is €56.75. Notice something uncomfortable: 9 of the 10 receipts are below the mean. That single €289 receipt has "pulled" the mean upwards. We will come back to this shortly, because it is the mean's main weakness.

• It uses all the information: every value influences the result.
• It is the balance point: the sum of the deviations from the mean is always zero (the data "weigh" the same on both sides).
• It lets you recover the total: if the average transaction value is €32.40 and there were 57,000 transactions across the chain yesterday, total sales were \( 32.40 \times 57{,}000 = 1{,}846{,}800 \) euros. Neither the median nor the mode allows this calculation; that is why finance works with means almost exclusively.
• It only makes sense for quantitative variables (and, by the convention seen in Types of Data, for ordinal variables like 0–10 satisfaction scores treated as interval data).

Sometimes not every value should carry the same weight. The weighted mean assigns each value \( x_i \) a weight \( w_i \):

\[ \bar{x}w = \frac{\sum{i=1}^{n} w_i , x_i}{\sum_{i=1}^{n} w_i} \]

Marta hands you the average transaction value of three stores and asks for "the combined average":

The tempting (and wrong) approach: simply average the three numbers:

\[ \frac{34.10 + 31.90 + 27.80}{3} = \frac{93.80}{3} \approx \text{€}31.27 \]

This treats Cuenca (420 transactions) as if it carried the same weight as Madrid-Centro (1,250 transactions). It does not represent the "typical" receipt of a random customer across the three stores.

The correct calculation (weighting by number of transactions):

\[ \bar{x}_w = \frac{34.10 \times 1{,}250 + 31.90 \times 980 + 27.80 \times 420}{1{,}250 + 980 + 420} = \frac{42{,}625 + 31{,}262 + 11{,}676}{2{,}650} = \frac{85{,}563}{2{,}650} \approx \text{€}32.29 \]

The difference (€32.29 versus €31.27) looks small, but applied to millions of transactions a year it translates into very different revenue forecasts. In fact, NovaMarket's benchmark average transaction value, €32.40, is a weighted mean across the 42 stores: the big stores weigh more because they generate more transactions.

Rule of thumb: whenever you average averages (of stores, channels, months), ask yourself whether the groups differ in size. If they do, weight.

The median is the value sitting in the middle position when the data are sorted from smallest to largest: it leaves 50% of the observations below it and 50% above.

Calculation:

1. Sort the data.
2. If \( n \) is odd, the median is the value in position \( \frac{n+1}{2} \).
3. If \( n \) is even, it is the mean of the two middle values (positions \( \frac{n}{2} \) and \( \frac{n}{2}+1 \)).

The data, already sorted (n = 10, even):

\[ 12.50 ;; 18.90 ;; 22.40 ;; 27.30 ;; \underbrace{29.80 ;; 31.60}_{\text{positions 5 and 6}} ;; 38.20 ;; 45.70 ;; 52.10 ;; 289.00 \]

\[ \text{Median} = \frac{29.80 + 31.60}{2} = \text{€}30.70 \]

Compare: mean €56.75, median €30.70. The median describes the "typical" receipt of that afternoon far better, because it is unaffected by whether the largest receipt is €289 or €2,890: it only counts how many values lie on each side, not how much the extremes are worth. The median is said to be a robust measure.

The median requires that the data can be ordered: it works for quantitative variables and also for ordinal ones (you can talk about the median satisfaction score), but not for nominal ones (there is no such thing as the "median payment method").

The mode is the value that occurs with the highest frequency. It is the only measure of central tendency that is valid for nominal variables.

Examples at NovaMarket:

• Among yesterday's transactions, the most frequent payment method was card (61% of transactions): that is the mode. No mean or median of the payment method is possible.
• Marta reviews a subsample of 15 responses from the satisfaction survey (0–10 scale):

The mode is 9 (5 responses). The median is position \( (15+1)/2 = 8 \) of the sorted data, which also lands on 9. The mean is \( 124/15 \approx 8.27 \). That all three sit so close together already hints at what the full 640 responses will confirm: a distribution concentrated in the high scores.

Notes on the mode:

• It may not exist (if no value repeats) or there may be several (bimodal distributions; you will see them graphically in Graphical Representation of Data).
• In continuous variables (exact amounts, times) hardly any value repeats, so the mode is computed on data grouped into intervals (the "modal interval") or simply not used.
• In business it is highly expressive: "the best-selling product", "the busiest time slot" are modes.

You have already seen it with the €289 receipt: the mean is sensitive to extreme values; the median is not. Let's verify it by removing that receipt:

A single value moved the mean by €25.81 (from 30.94 to 56.75) and the median by barely €0.90. In variables such as purchase amounts, salaries, wealth or waiting times — where a few enormous values are perfectly normal — the mean can paint a misleading picture of the typical case.

A compromise between the two is the trimmed mean: you discard a small percentage of the lowest and highest values (say, 10% at each end) and average the rest. With our 10 receipts, trimming 10% from each side means dropping the minimum (12.50) and the maximum (289.00):

\[ \bar{x}_{\text{trim}} = \frac{18.90 + 22.40 + 27.30 + 29.80 + 31.60 + 38.20 + 45.70 + 52.10}{8} = \frac{266.00}{8} = \text{€}33.25 \]

For now it is enough to know that it exists and what it is for: it is common in official indicators and in sports (judges' scores). In this course we will normally work with the mean and the median.

Two questions almost always settle the choice: what type of variable is it? and is there skewness or are there extreme values?

A very useful diagnostic hint: compare the mean and the median.

• Mean ≈ median → distribution roughly symmetric; the mean is safe.
• Mean clearly greater than the median → high values are pulling it up (right skew, typical of monetary amounts).
• Mean clearly less than the median → low values are pulling it down (left skew, typical of high satisfaction scores: the survey's mean of 8.1 will sit below its median).

And a professional tip: when the gap between the two is large, report both. "The mean transaction value was €56.75, but the median was €30.70: the mean is inflated by a few very large purchases" is a sentence that gives Marta more information than either number on its own.

There is one case where the arithmetic mean fails spectacularly: averaging rates of change that apply in a chain (year-on-year percentage growth, investment returns, and so on).

Sales in NovaMarket's online store evolved like this:

Arithmetic mean of the rates: \( (20 + 50 - 30)/3 = 13.3 \)% per year. But if sales had truly grown by 13.3% every year, in Year 3 they would be \( 1{,}000{,}000 \times 1.133^3 \approx \text{€}1{,}454{,}000 \)… and in reality they are €1,260,000. The arithmetic mean overestimates the growth, because percentages compound by multiplying, not by adding.

The correct measure is the geometric mean of the growth factors:

\[ \bar{x}_G = \sqrt[n]{x_1 \cdot x_2 \cdots x_n} \]

Step 1 — Multiply the factors: \( 1.20 \times 1.50 \times 0.70 = 1.26 \)

Step 2 — Take the n-th root (n = 3 years): \( \bar{x}_G = \sqrt[3]{1.26} \approx 1.080 \)

The true average annual growth is 8.0%. Check: \( 1{,}000{,}000 \times 1.080^3 \approx \text{€}1{,}259{,}700 \), which matches (up to rounding) the actual €1,260,000.

Rule: to average chained percentage changes, turn each rate into a factor (\( 1 + \text{rate} \)), take the geometric mean of the factors, and subtract 1 at the end.

• Reporting only the mean for skewed variables. With amounts, spending or times, always pair the mean with the median (or use the median outright). If mean ≫ median, say so explicitly.
• Averaging averages without weighting. The "average of the 42 stores' average transaction values" is not NovaMarket's average transaction value, unless every store handles the same number of transactions (they don't). Weight by size.
• Computing means of nominal variables coded as numbers. If "card = 1, cash = 2, mobile = 3", a mean of 1.7 means nothing. Numeric codes do not turn a nominal variable into a quantitative one (revisit Types of Data).
• Using the arithmetic mean with growth rates. For chained changes, use the geometric mean of the factors.
• Forgetting the context of the unit of analysis. "Average spend is €32.40" (per transaction) and "average spend is €187" (per customer per month) can both be true; always specify which unit you are averaging over.
• Tip: before computing anything, sort the data (even just mentally) and look at the extremes. Ten seconds spent on the maximum and minimum prevent a lot of misreadings.

A sample of 8 receipts from the Sevilla-Nervión store (in euros): 8.90 — 15.40 — 21.70 — 26.30 — 28.10 — 34.60 — 41.00 — 196.00.

1. Compute the mean and the median.
2. Which of the two would you report to Marta as the "typical receipt" of this sample, and why?
3. Recompute the mean without the €196.00 receipt. How much does it change? And the median?

The 640 responses to the satisfaction survey are split across the channel strata designed in the Data Collection and Sampling lesson: 448 from in-store customers, 128 from online customers and 64 from app customers. The mean satisfaction scores by channel are:

Compute the overall mean satisfaction of the 640 responses. Why would it be wrong to compute \( (8.3 + 7.6 + 7.9)/3 \)?

The number of Club Nova members changed by +15% in the first year, +5% in the second and −8% in the third.

1. Compute the average annual growth using the geometric mean.
2. Compare it with the arithmetic mean of the rates. Which one would you use to project membership into the future?

Exercise 1:

1. Sum: \( 8.90 + 15.40 + 21.70 + 26.30 + 28.10 + 34.60 + 41.00 + 196.00 = \text{€}372.00 \). Mean: \( 372.00 / 8 = \text{€}46.50 \). Median (n = 8, even; middle values in positions 4 and 5): \( (26.30 + 28.10)/2 = \text{€}27.20 \).
2. The median (€27.20): the mean (€46.50) sits above 7 of the 8 receipts, dragged up by the €196 purchase. As the "typical receipt" it is misleading. (Ideally, report both and explain the gap.)
3. Without the extreme value: \( 176.00 / 7 \approx \text{€}25.14 \). The mean drops by €21.36 (from 46.50 to 25.14). The new median is the 4th value of 7: €26.30 — it barely moves (€0.90). Common mistake: forgetting to re-sort and recount positions after removing a data point; the position of the median depends on \( n \).

Exercise 2:

\[ \bar{x}_w = \frac{8.3 \times 448 + 7.6 \times 128 + 7.9 \times 64}{640} = \frac{3{,}718.4 + 972.8 + 505.6}{640} = \frac{5{,}196.8}{640} \approx 8.1 \]

The overall mean is 8.1 (matching the survey's official figure). The unweighted mean, \( (8.3+7.6+7.9)/3 = 7.93 \), is wrong because it gives the 64 app respondents the same weight as the 448 in-store ones: it understates overall satisfaction. Common mistake: dividing by 3 (the number of groups) instead of 640 (the sum of the weights).

Exercise 3:

1. Factors: 1.15, 1.05 and 0.92. Product: \( 1.15 \times 1.05 \times 0.92 = 1.1109 \). Geometric mean: \( \sqrt[3]{1.1109} \approx 1.0357 \) → average annual growth ≈ 3.6%.
2. Arithmetic mean: \( (15 + 5 - 8)/3 = 4.0 \)%. For projections you must use the geometric mean (3.6%): it is the only constant rate that, applied over three years, reproduces the actual total growth (+11.1%). Common mistake: averaging the percentages directly (\( (0.15+0.05-0.08)/3 \)) instead of the factors; with negative rates the error gets worse.

You now know how to condense a data set into a central value and, more importantly, how to choose that value: the mean for totals and symmetric data, the weighted mean when groups carry different weights, the median when there is skewness or extreme values, the mode for nominal variables, and the geometric mean for chained growth. You have also picked up a professional reflex: comparing mean and median as a first diagnostic of the shape of the data.

But a central value, however well chosen, tells only half the story. Two NovaMarket stores can sell exactly the same on average and yet be, one of them, a stable and predictable business and, the other, a roller coaster that plays havoc with stock and staffing. To tell them apart you need to measure how far the data stray from their center: that is what Measures of Dispersion are for — the subject of the next lesson.