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Math 6 · S.ID.A.3

Mean as balance point

A list of numbers is spread along a line. Is there one point where it balances, and how does that point move when one number changes?

You need
The number lineGrades 1 to 2
Where you are 3 in Mean
Statistics7
Center and spread13
Mean3
Mean as balance point
Mean as fair share: computing the meanMean from a frequency table and weighted mean

Why this exists

Five friends compare their weekly pocket money: 4, 5, 5, 8 and 13 dollars. What amount is typical? One answer is the fair share: put all the money in one pile and share it out equally. The pile holds 35 dollars, and five equal shares are 7 dollars each. That number is the mean.

A calculation hides where its answer sits. Mark the five amounts on a number line and 7 lands in an odd spot: three friends below it, only two above, and nobody actually gets 7 dollars. Why there? This lesson gives the mean a place on the line: the point where the data balance, like a seesaw on its pivot. That picture explains why one large value drags the mean toward it, exactly how far the mean moves when one value changes, and how to recover a value that went missing when only the mean was kept.

The first day recaps the fair share and builds the balance picture. The second follows a changed value and finds a missing one.

Definition

The mean of a list of numbers is the total of the values divided by how many values there are:

\text{mean} = \frac{\text{total of the values}}{\text{number of values}}

It is the fair share: if the total were shared out so that every value became the same, each would be the mean. The full lesson on computing it is the mean as a fair share; the lines above are all this lesson needs.

A value’s distance from a point is how far apart the two sit on the number line: the larger number minus the smaller. Measured from the mean, a value below it is on the low side, a value above it on the high side, and a value equal to the mean has distance 0 and sits on neither side.

The balance point of a list is the point where the total of the distances on the low side equals the total of the distances on the high side. The main result below is that the balance point and the mean are the same point.

Intuition

Draw the data as a dot plot, and picture the number line as a long, weightless ruler with an identical coin taped at each dot. Put a pivot under the ruler. A coin far from the pivot tips the ruler harder than a coin close to it: twice as far out, twice the turning effect. So the ruler balances when the coins on the left, each counted by its distance from the pivot, add up to the same as the coins on the right. That is the balance point defined above.

3 4 5 6 7 8 9 10 11 12 13 14 3 2 2 1 6 mean 7 pocket money (dollars a week)
Figure 1The pocket money balances at its mean, 7. The distances on the left add to 7, and so do the distances on the right.

In Figure 1 the five amounts sit on the ruler with the pivot at 7. The low side has distances 3, 2 and 2, total 7; the high side has 1 and 6, total 7. Three coins close in on one side match two on the other, because one of the two is far out.

Moving coins shows why the fair share and the balance point must agree. Slide one coin up by some amount and another down by the same amount: the total does not change, so the fair share does not change. Start with 4, 5, 5, 8, 13, total 35. Move the 4 up by 3 and the 13 down by 3: 7, 5, 5, 8, 10. Move a 5 up by 2 and the 10 down by 2: 7, 7, 5, 8, 8. Move the last 5 up by 2 and each 8 down by 1: 7, 7, 7, 7, 7. Every move kept the total at 35 and took the same amount off the low side as off the high side, so the gap between the sides never changed. At the end all five coins are stacked at 7, the fair share, and both sides are 0. A gap of 0 at the end means a gap of 0 at the start: the ruler balanced at 7 all along.

Results

The mean is the balance point. For any list of numbers, the total distance of the values below the mean equals the total distance of the values above it. No other point has this property.

One changed value. If one value in a list of n values goes up by d, the mean goes up by \displaystyle \frac{d}{n}. If it goes down by d, the mean goes down by \displaystyle \frac{d}{n}.

One added value. A new value above the mean pulls the mean up, one below pulls it down, and one equal to the mean leaves it where it is. If the new value is d above the old mean and the list now has n values, the mean rises by \displaystyle \frac{d}{n} (and falls by \displaystyle \frac{d}{n} for a value d below).

A missing value. Since the mean is the total divided by the count,

\text{total} = \text{mean} \times \text{number of values},

and a missing value is that total minus the total of the values that are known.

Why it is true

Balance. Take a list of n values with mean m. Every high-side value is m plus its distance, every low-side value is m minus its distance, and a value at the mean is just m. Add all n values. Writing H for the total of the high-side distances and L for the low-side total,

\text{total} = n \times m + H - L.

But the mean is the total divided by n, so the total is exactly n \times m. The H and the L must cancel, and that means H = L.

The same bookkeeping works around any point p: the total is n \times p, plus the high-side distances from p, minus the low-side distances from p. If p balances, those two cancel, so the total is n \times p and \displaystyle p = \frac{\text{total}}{n}, which is the mean. No other point balances.

One changed value. Raising one value by d adds d to the total and leaves the count at n, so the new mean is \displaystyle \frac{\text{total} + d}{n} = \frac{\text{total}}{n} + \frac{d}{n}. Lowering it works the same way with d taken off.

One added value. If k values have mean m, their total is k \times m. Adding m + d makes the total (k + 1) \times m + d over k + 1 values, so the new mean is \displaystyle m + \frac{d}{k + 1}, and k + 1 is the new count n.

Worked examples

Examples 1 to 3 go with the first day, Examples 4 to 7 with the second.

Example 1

Fair share, then balance

Amadou, Anahi, Andrei, Anneli and Anouar get 4, 5, 5, 8 and 13 dollars of pocket money a week. Find the mean, then check that it is the balance point.

Solution

The total is 4 + 5 + 5 + 8 + 13 = 35 dollars, shared among 5 friends, so the mean is \displaystyle \frac{35}{5} = 7 dollars.

Measure every amount from 7. The low side holds 4, 5 and 5, at distances 3, 2 and 2, total 7. The high side holds 8 and 13, at distances 1 and 6, total 7. The totals are equal, so the plot balances at 7, as Figure 1 shows. The high side has fewer values, but Anouar’s 13 sits far out and its distance of 6 almost matches the whole low side on its own.

Example 2

Testing a guess

Six classmates measure their hand spans, thumb tip to little-finger tip with the hand stretched: 16, 17, 17, 19, 21 and 24 cm. One guess at the center is halfway between the smallest and largest, 20 cm. Is 20 the balance point? If not, which way does the balance point lie, and where is it?

Solution

Measure from 20. The low side holds 16, 17, 17 and 19, at distances 4, 3, 3 and 1, total 11. The high side holds 21 and 24, at distances 1 and 4, total 5. The low side is heavier, so a ruler on a pivot at 20 tips down on the left. To balance it the pivot has to slide left, toward the heavier side: the balance point is below 20.

The mean finds it exactly. The total is 16 + 17 + 17 + 19 + 21 + 24 = 114 over 6 values, so the mean is \displaystyle \frac{114}{6} = 19 cm. Check at 19: the low side has distances 3, 2 and 2, total 7; the 19 sits on the pivot with distance 0; the high side has 2 and 5, total 7. The halfway guess listened to two values; the balance point listens to all six.

Example 3

A mean that is not in the list

In one month, five classmates borrowed 2, 3, 3, 4 and 9 library books. Find the mean and check that the plot balances there.

Solution

The total is 21 books over 5 classmates, so the mean is \displaystyle \frac{21}{5} = 4.2 books. Nobody borrowed 4.2 books, and nobody could: the mean is a fair share, not a value someone must have.

Measured from 4.2, the low side holds 2, 3, 3 and 4, at distances 2.2, 1.2, 1.2 and 0.2, total 4.8. The high side holds only the 9, at distance 9 - 4.2 = 4.8. Balanced, as Figure 2 shows: one value far out outweighs four close in, and pulls the mean above four of the five values. How strongly a lone far value drags the mean is taken up in resistant and non resistant measures.

1 2 3 4 5 6 7 8 9 10 2.2 1.2 1.2 0.2 4.8 mean 4.2 books borrowed in a month
Figure 2Four values close to the mean on the left balance one value far out on the right. Both sides total 4.8.
Example 4

One value changes

Six cousins get 3, 5, 6, 6, 10 and 12 dollars of pocket money a week. One cousin’s allowance rises from 6 to 18 dollars. How does the mean change?

Solution

Before the change the total is 42, so the mean is \displaystyle \frac{42}{6} = 7 dollars. The raise adds 18 - 6 = 12 to the total and leaves the count at 6, so the new total is 54 and the new mean is \displaystyle \frac{54}{6} = 9 dollars.

The mean rose by 2, which is \displaystyle \frac{12}{6}: the raise of 12 is shared across all six cousins. The balance picture agrees. Measured from 9, the low side holds 3, 5 and 6, at distances 6, 4 and 3, total 13; the high side holds 10, 12 and 18, at distances 1, 3 and 9, total 13. Figure 3 shows the pivot sliding from 7 to 9. With the result \displaystyle \frac{d}{n} no adding is needed: \displaystyle 7 + \frac{12}{6} = 9.

before: mean 7 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 after: mean 9 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 pocket money (dollars a week)
Figure 3Raising one of six values by 12 moves the balance point up by \displaystyle \frac{12}{6} = 2.
Example 5

Adding a value

Four friends have hand spans of 16, 17, 19 and 20 cm. A fifth friend with a span of 23 cm joins them. What is the new mean? What fifth span would leave the mean unchanged, and what fifth span would lower it to 17 cm?

Solution

The four spans total 72, so their mean is \displaystyle \frac{72}{4} = 18 cm. With the fifth the total is 95 over 5 values, and the mean is \displaystyle \frac{95}{5} = 19 cm. The new span is 23 - 18 = 5 above the old mean, and that 5 is shared among the 5 values now in the list: a rise of \displaystyle \frac{5}{5} = 1.

A fifth span of 18 cm leaves the mean alone, since \displaystyle \frac{90}{5} = 18. On the ruler, a coin placed on the pivot has distance 0 and tips neither side.

To lower the mean to 17, the five spans must total 5 \times 17 = 85, so the fifth is 85 - 72 = 13 cm. It sits 5 below the old mean, and \displaystyle \frac{5}{5} = 1 is the drop.

Example 6

A missing value

Last term five classmates each counted the library books they borrowed. The mean was 6 books, but one count was lost; the other four were 4, 5, 7 and 9. Find the lost count in two ways.

Solution

By the total: five counts with mean 6 total 5 \times 6 = 30. The four known counts total 25, so the lost one is 30 - 25 = 5.

By the balance: the plot must balance at 6. Of the known counts, the low side holds 4 and 5, at distances 2 and 1, total 3; the high side holds 7 and 9, at distances 1 and 3, total 4. The high side is heavier by 1, so the lost count must add 1 to the low side: it is 6 - 1 = 5. Both ways agree.

Example 7

Raising a mean with one more value

Anouar’s pocket money over five weeks has a mean of 8 dollars. How much must he get in the sixth week for the mean of all six weeks to be 9 dollars?

Solution

By totals: five weeks with mean 8 total 5 \times 8 = 40 dollars. Six weeks with mean 9 must total 6 \times 9 = 54. The sixth week must bring 54 - 40 = 14 dollars.

By the added-value result, run backwards: a new value d above the old mean raises the mean by \displaystyle \frac{d}{6}, since the list will have 6 values. A rise of 1 needs \displaystyle \frac{d}{6} = 1, so d = 6, and the sixth week is 8 + 6 = 14 dollars. Neither way needed the five separate amounts: the mean and the count carry the total, the idea behind the mean from a frequency table.

Boundaries

  • The mean and its balance need numbers on a number line. Categorical data, such as favorite colors, has no mean (see types of data).
  • The balance is of distances, not of counts: Example 3 balances four values against one. The center that splits the count in half is the median, and on lopsided data the two differ.
  • The rule \displaystyle \frac{d}{n} for a changed value holds only when the count stays the same. Adding or removing a value changes n as well, so go back to totals or use the added-value result.
  • The mean need not be one of the values, or even a possible value: 4.2 books.

Where this goes next

The distances measured here are the raw material of the mean absolute deviation, which averages them to say how spread out a list is. With a spread in hand, two groups can be compared by how far apart their balance points sit, in comparing two populations. Because a lone far value drags the balance point while the median barely moves, the two centers separate on lopsided data, which is how choosing a center for skewed data tells them apart.

Exercises

Exercises 1 to 13 go with the first day (Examples 1 to 3), and Exercises 14 to 25 with the second (Examples 4 to 7).

Day 1: the mean and its balance

  1. Four friends get 3, 4, 8, 9 dollars of pocket money a week. Find the mean and the total distance on each side of it.
  2. Five hand spans measure 14, 15, 15, 16, 20 cm. Find the mean and the total distance on each side of it.
  3. Find the mean of 1.5, 2, 3.5, 5 and check that it is the balance point.
  4. Six classmates borrowed 0, 2, 3, 3, 4, 6 library books in a week. Find the mean and check that it is the balance point.

Day 1: testing a point

  1. Is 10 the balance point of 6, 9, 11, 16? If not, which way does the balance point lie? Find it.
  2. In the list 2, 4, 4, 5, 10, more values sit at 4 than anywhere else. Is 4 the balance point? Find the balance point.
  3. For 1, 2, 9, test the point halfway between the smallest and largest value. Which side is heavier? Find the mean and check it.
  4. Pocket money amounts of 2, 6, 7, 9 dollars balance at one of 5, 6 or 7. Test each, then confirm with the fair share.

Day 1: placing a value

  1. A dot plot shows the values 5 and 12. Where must a third value go for the plot to balance at 8?
  2. A dot plot shows 10, 11, 15, 20. Where must a fifth value go for the plot to balance at 14?

Day 1: reasoning

  1. Explain why the mean of a list can never be smaller than every value in the list.
  2. The list 1, 1, 2, 2, 9 has one value far out on its own. Find the mean, and explain with distances how one value balances four.
  3. In the list 3, 5, 10, the 3 is moved up by 2 and the 10 is moved down by 2. Explain why the mean does not change, then check by finding both means.

Day 2: one changed value

  1. Six values have a mean of 10. One of them is raised by 9. What is the new mean?
  2. Four friends get 5, 6, 8, 9 dollars a week. The one who got 9 dollars now gets 5. Find the old and new means, and the change.
  3. Eight hand spans have a mean of 17 cm. One was written down as 15 cm but was really 19 cm. What is the correct mean?
  4. In a list of 5 values, by how much must one value rise to raise the mean by 3?

Day 2: adding or removing a value

  1. A class borrowed a mean of 12 library books a week over 4 weeks. Find the mean over 5 weeks if the fifth week’s count is (a) 22, (b) 12, (c) 2.
  2. The list 3, 4, 6, 8, 9 has mean 6. Remove the 9. What is the new mean, and why did it go down?

Day 2: missing values

  1. Six hand spans have a mean of 18 cm. Five of them are 16, 17, 18, 19, 21 cm. Find the sixth.
  2. Four weeks of pocket money have a mean of 8 dollars. Three of the weeks were 6, 9, 7 dollars. Find the fourth week in two ways: with the total, and with the balance.
  3. Three weeks of pocket money have a mean of 5 dollars. How much in the fourth week would make the mean of all four weeks 6 dollars?

Day 2: reasoning

  1. Explain in two ways, with totals and with the balance, why adding a value equal to the mean leaves the mean unchanged.
  2. Two values in a list of 8 each rise by 4. Explain why the mean rises by 1.
  3. A new value is added to a list. Explain why the new mean always lies between the old mean and the new value, unless the new value equals the old mean, when nothing moves.
Answers
  1. Mean 6 dollars: \displaystyle \frac{24}{4} = 6. Low side 3 + 2 = 5, high side 2 + 3 = 5.
  2. Mean 16 cm: \displaystyle \frac{80}{5} = 16. Low side 2 + 1 + 1 = 4, high side 4 from the 20; the 16 sits on the balance point.
  3. Mean 3: \displaystyle \frac{12}{4} = 3. Low side 1.5 + 1 = 2.5, high side 0.5 + 2 = 2.5. Balanced.
  4. Mean 3 books: \displaystyle \frac{18}{6} = 3. Low side 3 + 1 = 4, high side 1 + 3 = 4; the two 3s sit on the point.
  5. No; the balance point is 10.5. At 10 the low side totals 4 + 1 = 5 and the high side 1 + 6 = 7, so the balance point lies above 10. The mean is \displaystyle \frac{42}{4} = 10.5, and there each side totals 6: 4.5 + 1.5 and 0.5 + 5.5.
  6. No; the balance point is 5. At 4 the low side totals 2 and the high side 1 + 6 = 7. The mean is \displaystyle \frac{25}{5} = 5; there the low side is 3 + 1 + 1 = 5 and the high side 5.
  7. Halfway is 5. There the low side totals 4 + 3 = 7 and the high side 4, so the low side is heavier and the mean is below 5. The mean is \displaystyle \frac{12}{3} = 4; at 4 each side totals 5.
  8. The balance point is 6. At 5: low 3, high 1 + 2 + 4 = 7. At 6: low 4, high 1 + 3 = 4. At 7: low 5 + 1 = 6, high 2. Fair share: \displaystyle \frac{24}{4} = 6 dollars.
  9. At 7. The 5 puts 3 on the low side and the 12 puts 4 on the high side, so the third value must add 1 to the low side: 8 - 1 = 7. Check: \displaystyle \frac{24}{3} = 8.
  10. At 14. The low side already totals 4 + 3 = 7 and the high side 1 + 6 = 7, so the fifth value must add nothing to either side. Check: \displaystyle \frac{70}{5} = 14.
  11. If the mean were smaller than every value, every value would be on the high side, which would total more than 0 while the low side totaled 0. The mean always balances, so this cannot happen. (If all the values are equal, they sit on the mean, which is then not smaller.)
  12. Mean 3: \displaystyle \frac{15}{5} = 3. Low side 2 + 2 + 1 + 1 = 6, high side 9 - 3 = 6. Each low value sits close to the mean and adds a little; the 9 sits far out, and its one distance matches their four together.
  13. Moving the 3 up by 2 adds 2 to the total and moving the 10 down by 2 takes 2 away, so the total and count stay the same, and so does the fair share. Both means are 6: \displaystyle \frac{18}{3} = 6 for 3, 5, 10 and for 5, 5, 8.
  14. 11.5. The total rises by 9, shared among 6 values: \displaystyle 10 + \frac{9}{6} = 11.5.
  15. Old mean 7 dollars, new mean 6 dollars, a drop of 1. The totals are 28 and 24; the drop of 4 shared among 4 friends is \displaystyle \frac{4}{4} = 1.
  16. 17.5 cm. Correcting the value adds 19 - 15 = 4 to the total: \displaystyle 17 + \frac{4}{8} = 17.5.
  17. By 15. The mean rises by \displaystyle \frac{d}{5}, and \displaystyle \frac{d}{5} = 3 when d = 15.
  18. (a) 14, (b) 12, (c) 10. The four weeks total 4 \times 12 = 48, so the means are \displaystyle \frac{70}{5} = 14, \displaystyle \frac{60}{5} = 12 and \displaystyle \frac{50}{5} = 10: each new count moves the mean by its distance from 12 divided by 5.
  19. 5.25. The total drops to 30 - 9 = 21 over 4 values: \displaystyle \frac{21}{4} = 5.25. The 9 was on the high side; without it the low side is heavier, so the balance point slides down.
  20. 17 cm. The six spans total 6 \times 18 = 108, the five known total 91, and 108 - 91 = 17.
  21. 10 dollars. By the total: 4 \times 8 = 32 and 32 - 22 = 10. By the balance at 8: the low side holds 6 and 7, total 2 + 1 = 3; the high side holds 9, total 1; so the fourth week must add 2 to the high side: 8 + 2 = 10.
  22. 9 dollars. Four weeks need 4 \times 6 = 24, three weeks give 3 \times 5 = 15, and 24 - 15 = 9.
  23. With totals: k values with mean m total k \times m; adding m makes (k + 1) \times m over k + 1 values, so the mean is still m. With the balance: the new value sits on the pivot at distance 0 and adds nothing to either side, so the plot still balances at m.
  24. The total rises by 4 + 4 = 8 while the count stays 8, so the mean rises by \displaystyle \frac{8}{8} = 1.
  25. Say the new value is d above the old mean and the list now has n values, at least 2. The mean rises by \displaystyle \frac{d}{n}, more than 0 but at most half of d, so the new mean is above the old mean and below the new value. A value below the old mean works the same way downwards, and if d = 0 nothing moves.
kg1128 · updated 2026-10-13 · CC-BY-SA-4.0