Mean as a balance point

Imagine placing the numbers $3$ and $7$ as dots on a number line and resting the line on a single point so it doesn't tip. Where would you put the point? Right in the middle, at $5$. That is exactly the average , and that little see-saw is the secret of what we now call the arithmetic mean.

Concept

You already know the rule: for $n$ values $x_1, x_2, \dots, x_n$, the mean is

$\bar{x} = \dfrac{x_1 + x_2 + \dots + x_n}{n}.$

That is the computational definition. There is also a physical definition that gives much more intuition: the mean is the balance point of the data on a number line.

What does "balance" mean here? Imagine the number line is a thin ruler, and each data value is a small weight placed at its position. The mean is the point at which you can support the ruler with one finger and it won't tip to either side. Mathematically, this comes down to a beautiful identity:

$\text{(sum of distances from mean, on the left)} = \text{(sum of distances from mean, on the right)}.$

For two numbers it is obvious , the balance point is halfway between them. But the same idea works for many values, and that is the surprise.

Example. Data: $6, 6, 8, 8$. Mean: $\bar{x} = \dfrac{6+6+8+8}{4} = 7$.

• LHS distances: $7-6=1$ and $7-6=1$. Total $= 2$.
• RHS distances: $8-7=1$ and $8-7=1$. Total $= 2$.

They match. Now try with three values clustered together and one far away.

Example. Data: $9, 11, 11, 17$. Mean $= \dfrac{9+11+11+17}{4} = 12$.

• LHS distances from $12$: $3, 1, 1$. Sum $= 5$.
• RHS distance: $5$. Sum $= 5$.

Again they match , but notice that the mean ($12$) is not the midpoint of the smallest and largest values ($\dfrac{9+17}{2} = 13$). The mean depends on where every point is, not just the extremes.

A key consequence. If you know the mean $\bar{x}$ and the number of values $n$, then the total is forced:

$x_1 + x_2 + \dots + x_n = n \cdot \bar{x}.$

Adding a new value pulls the mean towards the new value. Adding a value equal to the current mean doesn't move the mean at all , you have placed weight exactly on the pivot. Adding a value much larger than the mean shifts the balance point to the right; a value much smaller shifts it left. This is why a single extremely high or low number , an outlier , can drag the mean far from where most of the data sits.

Worked examples

Example 1. Find the mean of $4, 6, 8, 10, 12$.

• Sum $= 40$. Count $= 5$. Mean $= 40/5 = 8$.
• Check the balance: LHS distances $4, 2$; RHS distances $2, 4$. Sums both equal $6$. ✓

Example 2. The mean of five test scores is $80$. The first four are $75, 82, 78, 85$. Find the fifth.

• Total marks $= 5 \times 80 = 400$.
• Sum of first four $= 75 + 82 + 78 + 85 = 320$.
• Fifth $= 400 - 320 = 80$.

Example 3. A teacher said the mean height of the $24$ students is $150.2$ cm, but then realises everyone was wearing shoes that add $1$ cm. What is the correct mean?

• Each measurement is $1$ cm too big, so the corrected mean is $150.2 - 1 = 149.2$ cm. No remeasuring needed!

Example 4. The dot plot shows values $4, 7, 8, 8, 9, ?$ with mean $9$. Find the missing value.

• Sum needed $= 6 \times 9 = 54$. Sum of known values $= 4+7+8+8+9 = 36$. Missing value $= 54 - 36 = 18$.

Try it yourself

1. Find the mean of $2, 5, 5, 8, 10$. Verify the balance using LHS and RHS distances.
2. Mean of $7$ numbers is $12$. What is their total?
3. If $\{10, 12, 14, x\}$ has mean $13$, find $x$.
4. The mean of $8, 13, 10, 4, 5, 20, y, 10$ is $10.375$. Find $y$.
5. A class of $15$ has mean weight $40$ kg. A new student of weight $48$ kg joins. New mean?
6. Mean of $9, 11, 11, 17$ is $12$. Is the midpoint of the data range also $12$? Explain.
7. A set has mean $50$. You add the value $50$ to it. Does the mean change? Why?
8. Two values $a$ and $b$ have mean $20$. If $a$ increases by $4$, by how much must $b$ decrease to keep the mean the same?

Activity

Pencil balance. Draw a number line on a strip of cardboard and mark eight equally spaced ticks $1$ to $8$. Place small coins (same value) on, say, ticks $2, 3, 5, 7$. Compute the mean ($\bar{x} = 4.25$). Now try to balance the strip on a pencil placed under the $4.25$ mark. With a bit of patience it should sit level. Move the pencil left or right and feel it tip. The arithmetic mean is literally where the data balances.