One number for the whole pile — the mean balances it, the median holds the middle, the mode names the favorite.
A pile of numbers is hard to hold in the mind all at once. The first thing a statistician does is squeeze the pile down to one number that stands for the whole — its center. But "center" is not one idea. You can balance the data, you can find its middle, or you can name its most common value, and these three can disagree wildly. This lesson builds all three — the mean, the median, and the mode — shows when each one tells the truth and when one billionaire can make the mean lie, and ends with the weighted mean, the average that lets some numbers count more than others.
Pool everyone's money into one pot and split it evenly — how much does each person get? That equal share is the mean, the everyday "average." Add up all n values and divide by how many there are:
x̄ = x₁ + x₂ + · · · + xₙn = Σ xᵢn
The bar over the x is read "x-bar." The numerator is the total of every data value; the denominator is the count n. Mechanically, the mean is the balance point: imagine each value as a one-gram weight sitting on a ruler. The mean is the single spot where the ruler would tip neither way — the deviations to the left exactly cancel the deviations to the right.
Five quiz scores: 3, 5, 5, 6, 8. Add them, then divide by 5:
x̄ = (3 + 5 + 5 + 6 + 8) / 5 = 27 / 5 = 5.4.
No score is actually 5.4 — the mean need not be a real data value. It is the fair share: if all 27 points were redistributed equally, everyone would hold 5.4.
Below, the dots are your data and the green triangle is the live mean. Drag the right-most value and watch the balance point chase it.
Line everyone up shortest to tallest and walk to the person standing dead center — that height is the median. It cares about position, not amount. Sort the data; the median is the middle value when n is odd, and the average of the two middle values when n is even.
Odd n. Sort 8, 3, 6, 5, 5 → 3, 5, 5, 6, 8. Five values, so the 3rd is the middle: median = 5.
Even n. Sort 4, 7, 9, 12. The two middles are 7 and 9, so median = (7 + 9) / 2 = 8 — a value that is not in the list, just like a mean can fail to be.
Because the median only asks "which value is in the middle slot," changing the size of an extreme value does nothing to it. Push the top score from 8 to 800 and the middle person has not moved — they are still the middle person. That stubbornness is exactly what makes the median robust.
The mean listens to every value's size, so a single huge value can yank it. The median listens only to the order, so it shrugs the huge value off.
The mode is the value that shows up most often — the best-selling shoe size, the most common quiz score, the answer a store actually stocks. Unlike the mean and median, the mode works on data that has no order at all (favorite colors, blood types), and there can be more than one of it.
Now try all three centers at once. Toggle a lone outlier in or out and watch which statistics flinch.
Picture a small shop where the five customers in line carry $24, $26, $28, $30, $32. Both the mean and the median equal $28 — for tidy, symmetric data, the two agree. Now a billionaire joins the line with $1000 in pocket cash:
| line | 24, 26, 28, 30, 32 | 24, 26, 28, 30, 32, 1000 |
|---|---|---|
| mean x̄ | 28 | 190 |
| median | 28 | 29 |
The $1000 dragged the mean from 28 all the way to 190 — a "typical" customer who is richer than every ordinary shopper in the line. That is absurd, and it is the mean's fatal flaw: it is not robust. The median only crept from 28 to 29, because adding a sixth person just shifted the middle slot by half a step. When data is skewed or carries outliers, the median is the more honest center; for symmetric data, mean ≈ median and either is fine.
One outlier can drag the mean far from the crowd, so "the average American household has $X in savings" (a mean) can be hugely inflated by a few billionaires; the median household tells a truer story. And beware the word "average" itself — it can mean the mean or the median, and a headline that does not say which is hiding something.
One more trap: never average the three measures together. The mean, median, and mode are three different answers to "where is the center," not three estimates of one number to be blended.
So which do you reach for? Roughly: report the mean for symmetric data with no wild values (test scores, repeated measurements); report the median for skewed data with outliers (incomes, house prices, response times); report the mode for categories or "what's most common" questions (the size to stock, the most-chosen option).
In a course the final exam usually counts more than a single homework. A plain mean treats every score equally — but that is not how the grade book works. The weighted mean attaches a weight wᵢ to each value, multiplies, sums, and divides by the total weight:
x̄ = w₁x₁ + w₂x₂ + · · · + wₙxₙw₁ + w₂ + · · · + wₙ = Σ wᵢ xᵢΣ wᵢ
If every weight is equal, the weights cancel and you get back the plain mean — so the ordinary mean is just the weighted mean with all weights the same.
Three scores 90, 80, 70 with weights 1, 1, 2 (the last counts double). Multiply, add, divide by the total weight 1 + 1 + 2 = 4:
x̄ = (1·90 + 1·80 + 2·70) / 4 = (90 + 80 + 140) / 4 = 310 / 4 = 77.5.
The plain mean would be (90 + 80 + 70)/3 = 80. The heavy weight on the lowest score pulled the weighted mean down to 77.5.
Set the marks and slide the weights below. Notice the weighted mean drifts toward whichever score carries the most weight.
| measure | what it is | formula | robust to an outlier? |
|---|---|---|---|
| mean x̄ | the balance point / fair share | Σxᵢ / n | no — gets dragged |
| median | the middle value in order | middle (or mean of two middles) | yes — barely moves |
| mode | the most frequent value | highest count (0, 1, or more) | yes — count-based |
| weighted mean | some values count more | Σwᵢxᵢ / Σwᵢ | no — like the mean |
One number for the center is never the whole story — two data sets can share the same mean yet feel completely different, one calm and one wild. That second question, how spread out is the data?, is the job of the next lesson: Quartiles & Dispersion. (And the symmetric "mean ≈ median" pile-up is the bell-shaped normal curve from Statistical Inference.)
Find the mean and the median of 4, 8, 8, 10, 15.
Mean = (4 + 8 + 8 + 10 + 15)/5 = 45/5 = 9. Sorted, the middle (3rd) value is 8, so the median = 8. The mean sits a touch above the median because the 15 pulls it up.
A data set is 2, 5, 7, 10 (an even count). Find the median.
With n = 4 the two middle values are 5 and 7. The median = (5 + 7)/2 = 6 — a value not in the list, which is fine for a median.
The seven test scores in a class are 70, 72, 72, 75, 78, 80, 95. Give the mode, and say whether the mean or the median is the better summary here.
72 appears twice and everything else once, so the mode = 72. The lone 95 is a high outlier, so it pulls the mean (≈ 77.4) above the median = 75. The median is the more honest summary of a "typical" student.
A student earns 80 on homework (weight 20%), 72 on the midterm (weight 30%), and 90 on the final (weight 50%). What is the weighted course grade?
x̄ = (0.20·80 + 0.30·72 + 0.50·90) / (0.20 + 0.30 + 0.50) = (16 + 21.6 + 45)/1 = 82.6. The heavy final lifts the grade above the plain average of (80+72+90)/3 ≈ 80.7.
Four numbers have a mean of 12. A fifth number, 22, is added. What is the new mean?
A mean of 12 over 4 values means the total is 4·12 = 48. Add 22: the new total is 70 over 5 values, so the new mean = 70/5 = 14. (Working backward from "mean × count = total" is the key move.)
A news headline says "the average salary at TechCo is $180,000." The 10 employees actually earn nine salaries of $60,000 and one CEO salary of $1,260,000. Is the headline misleading? What single number would you report?
The headline used the mean: (9·60,000 + 1,260,000)/10 = 1,800,000/10 = $180,000 — but nine of ten workers earn far less. The lone CEO salary is an outlier dragging the mean up. The median is $60,000 (the middle of ten $60k earners is the average of the 5th and 6th, both 60,000), a far truer "typical" pay. Report the median.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the three measures of center — mean, median, and mode — plus the weighted mean, addressing CCSS HSS-ID.A.2 (use statistics appropriate to the shape of the data to compare the center of two or more data sets) and HSS-ID.A.3 (interpret differences in center, accounting for the effect of outliers). Every statistic on this page — each mean, median, mode, and weighted mean — is computed by the shared GX helpers and never read off a picture, so the balance-point fulcrum and median tick always match the arithmetic. A natural extension is the next lesson on spread (HSS-ID.A.1/2/3: box plots, IQR, and standard deviation), where center alone is shown to be insufficient.