Skip to lesson content

Lesson 1 of 13

Tales by Dots and Lines · Lesson 1 of 13

The Mean as a Balance Point

“Understand the mean through fair sharing, dot plots, and balanced distances.”

Learning Objectives

• Calculate and interpret the arithmetic mean. • Locate the mean on a dot plot and compare distances on both sides. • Explain why the midpoint of the extreme values need not be the mean. • Connect balancing with fair sharing.

Four friends collect 2, 4, 6, and 8 smooth stones. If they put the stones together and share them equally, each friend receives 5. The number 5 summarises their collection, although nobody originally had exactly 5 stones. You may already know this fair-share meaning of the mean.

Now imagine placing one identical counter above each collection size on a number line. Could one point describe where all these counters balance? Exploring that question helps us understand why we add every value and then divide by the number of values. The mean becomes a location we can explain, not just a calculation to remember.

The Balancing Act

Start with two values, 4 and 10. Their mean is 7. On a number line, 7 lies 3 units from each value, so the distances on its left and right match. For two equally weighted observations, the mean is exactly their midpoint.

With more values, we compare total distances rather than the number of dots. A single dot far from the mean can balance several dots close to it. Repeated values still count separately: three dots at 4 represent three observations, each contributing its own distance.

Definition
Arithmetic mean

The sum of all observations divided by the number of observations. Each observation is included, even when several observations have the same value.

Mean as a fair shareLaTeX
Use the same unit for every observation. The mean has that unit too.
Equal total distances on both sidesEqual total distances on both sides123456789Mean = 5
Equal total distances on both sides— Left distances: 3 + 1 = 4. Right distances: 1 + 3 = 4.
Balancing two measurements

Problem
Two ribbons are 7 cm and 13 cm long. Find their mean length and check the balance.

  1. 1.Add the two lengths: 7 + 13 = 20 cm.
  2. 2.Divide by 2 ribbons: 20 ÷ 2 = 10 cm.
  3. 3.The left distance is 10 − 7 = 3 cm and the right distance is 13 − 10 = 3 cm.
  4. 4.The equal distances confirm that 10 cm is the balance point.

Balancing Several Dots

For the values 2, 3, and 10, the total is 15 and the mean is 5. Two observations lie below 5 and one lies above it. The distances below are 3 and 2, making 5 altogether. The distance above is also 5. Equal numbers of dots on each side are therefore unnecessary.

A value exactly at the mean contributes zero distance. It does not upset the balance. This observation will help later when we add a new value equal to an existing mean. The value joins the data, but it creates neither a shortage nor an excess relative to the balance point.

More dots can be on one sideMore dots can be on one side1234567891011Mean = 5
More dots can be on one side— Two left-hand dots balance one right-hand dot because distance matters.
Checking a proposed mean

Problem
A learner says that 6 is the mean of 2, 3, and 10. Use both arithmetic and distances to investigate.

  1. 1.The total is 2 + 3 + 10 = 15, and 15 ÷ 3 = 5.
  2. 2.At the proposed point 6, the left distances are 4 + 3 = 7, while the right distance is 4. These do not balance.
  3. 3.At 5, the left distances are 3 + 2 = 5 and the right distance is 5.
  4. 4.The correct mean is 5. Counting dots alone would not identify it.

Why There Is Only One Balance Point

Imagine temporarily giving every observation the value of a proposed centre. If the centre is too high, the total of these equal shares is larger than the actual total. If the centre is too low, the total of the shares is smaller. Exactly one value makes the two totals equal: the actual total divided by the count.

Another way to say this is that signed differences from the mean add to zero. Values below the mean give negative differences; values above it give positive differences. The positive and negative amounts cancel. You can use this idea to check a calculation without physically balancing anything.

Using excesses and shortages

Problem
Five jars hold 3, 4, 5, 5, and 8 marbles. Explain why the mean is 5.

  1. 1.The total is 25 marbles and there are 5 jars, so the mean is 25 ÷ 5 = 5.
  2. 2.Compared with 5, the first two jars are short by 2 and 1 marbles. The next two jars have no shortage.
  3. 3.The last jar has 3 extra marbles. Its excess matches the total shortage, 2 + 1 = 3.
  4. 4.Moving 2 marbles to the first jar and 1 to the second produces five equal shares.

The Middle of the Extremes Can Mislead

The smallest and largest values describe the ends of a dataset. Their midpoint uses only those two observations. The mean uses every observation, so it is influenced by how the data is distributed between the ends. The midpoint of the extremes and the mean agree for some datasets, but you must not assume they agree for all datasets.

Same extremes, different mean

Problem
Compare the datasets 2, 4, 6, 8 and 2, 2, 2, 8.

  1. 1.Both datasets have minimum 2 and maximum 8. Their extreme-value midpoint is (2 + 8) ÷ 2 = 5.
  2. 2.The first dataset has total 20, so its mean is 5.
  3. 3.The second has total 14, so its mean is 14 ÷ 4 = 3.5.
  4. 4.The extra small observations pull the second mean towards 2, even though its extreme values are unchanged.

Try a Fair-Share Model

Use small objects or sketches to connect a calculation with a physical model.

  1. Draw five piles containing 1, 3, 4, 6, and 6 objects.
  2. Calculate the total and predict the equal share before moving anything.
  3. Move objects from larger piles to smaller piles, keeping the total fixed.
  4. Plot the original values and check total distances from the equal share.

The total is 20 and the equal share is 4. Left distances are 3 + 1 = 4; right distances are 2 + 2 = 4. The pile already at 4 contributes zero distance.

Watch out

The mean can be a value that nobody recorded, and it can be fractional even when all the observations are whole numbers. Include every repeated observation in the count and total.

Quiz

Quick check

What is the mean of 3, 5, and 10?

Quick check

At the mean, what balances?

Quick check

Which dataset has mean 4?

Quick check

For 1, 1, and 10, is the mean the midpoint of 1 and 10?

Quick check

A data value equals the mean. What distance does it contribute?

The total is 18. There are 3 observations, so the mean is 18 ÷ 3 = 6.

Practice Problems

Practice Problems
  1. Find the mean of 6, 8, 10, and 12. Check the distances.
  2. Find the mean of 1, 2, 2, and 11. Explain why three dots can balance one.
  3. A learner averages only 2 and 12 from the data 2, 2, 4, 12. Correct the method.
  4. Construct four whole-number values with mean 6, none equal to 6.
  5. Could a mean be smaller than every observation? Explain.
  6. Two containers hold 4 L and 9 L. Find the fair share without losing any water.

Step 1: The total is 36 and the count is 4, giving mean 9. Step 2: Left distances are 3 + 1 = 4; right distances are 1 + 3 = 4.

Key Takeaways

Key Takeaways

• The mean uses the total and the number of all observations. • At the mean, total distances to the left and right are equal. • The mean need not be an observed value or a whole number. • The midpoint of the extreme values is not generally the mean.