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Grade 5 · 6.SP.B.5C

Mean as fair share: computing the mean

Three friends have stacks of cards of different heights. If they pooled their cards and shared them out evenly, how many would each one get, and what does that one number say about all of them?

You need
What division meansGrade 3
Where you are 3 in Mean
Statistics7
Center and spread13
Mean3
Mean as balance point
Mean as fair share: computing the mean
Mean from a frequency table and weighted mean

Why this exists

Chiamaka, Ilker and Noelani each keep a stack of trading cards. Chiamaka has 7, Ilker has 2 and Noelani has 6. Who has “about how many”? No single stack answers that. Chiamaka’s is the biggest, Ilker’s the smallest, and neither speaks for the group.

Here is a question that does have one answer: if the three friends put all their cards in one pile and shared the pile out evenly, how many would each get? That number uses every stack, and it is the same for everyone. It is called the mean of the numbers.

The mean lets you describe a whole list with one number. It lets you compare two lists of different lengths, like five days of cookies against seven. And because it is tied to the total, it lets you work backward: if you know the mean and how many values there are, you know the total, and from the total you can find a value that went missing.

The lesson takes three days. Part 1 levels towers and finds the mean as the total divided by how many. Part 2 meets means that are not whole numbers, including means of fractions. Part 3 works backward from the mean to a missing value, and asks what happens to the mean when a value changes or a new one is added.

Definition

The mean of a list of numbers is the amount each one would be if the total were shared out evenly among them.

Sharing a total evenly into a number of equal parts is division (division as sharing: the total is split into equal groups, one group for each). So the mean is found by adding all the values and dividing by how many values there are:

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

The values can be counts (cards, cookies) or measurements in one unit (cups, inches, dollars). They can be whole numbers, fractions or decimals. Every value counts, including a value of 0. The mean is also called the average. The parent star Mean gathers the different ways of looking at this number.

Intuition

Part 1: leveling the towers

Build each stack of cards as a tower of cubes, one cube per card. Chiamaka’s tower has 7 cubes, Ilker’s 2 and Noelani’s 6. Now level them: take cubes off the tall towers and put them on the short one, until all the towers are the same height.

7 2 6 5 5 5 before after
Figure 1Chiamaka, Ilker and Noelani level their stacks of 7, 2 and 6 cards. The 3 shaded cubes fill the 3 dashed spaces exactly, and every tower ends at 5.

Chiamaka gives 2 cubes and Noelani gives 1. Ilker takes all 3. Then each tower has 5 cubes. No cube was made and none was thrown away, so the three towers still hold 7 + 2 + 6 = 15 cubes, now in 3 equal towers. That is why the level height is 15 \div 3 = 5.

Two things are always true when you level.

  • What is given equals what is taken. Every cube that leaves a tall tower lands on a short one. Here 2 + 1 given matches 3 taken.
  • The level is never above the tallest or below the shortest. The tallest tower only ever gives and the shortest only ever takes, so the level sits between them (or at them, if every tower was already the same height).

You do not have to move cubes one at a time to find the level. Adding finds the total in one step, and dividing shares it in one step. The towers are the meaning; the division is the fast way.

Part 2: when the share is not whole

Sometimes the total does not split evenly. Suppose 17 cookies are shared among 3 friends. Each gets 5 whole cookies, which uses 15, and 2 are left. Cookies can be cut, so cut each leftover cookie into 3 equal pieces. That makes 6 thirds, and each friend gets 2 of them.

C I N C I N
Figure 2After 5 whole cookies each, 2 cookies are left. Cut into thirds, they give each friend \displaystyle \frac{2}{3} of a cookie more.

So each friend’s fair share is \displaystyle 5\frac{2}{3} cookies, and that is the mean. It is the same as the fraction \displaystyle \frac{17}{3}, since a fraction is a division (What division means).

A mean like \displaystyle 5\frac{2}{3} is not one of the numbers in the list, and it may not be an amount anyone could really hold. With cards, which cannot be cut, a mean of \displaystyle 5\frac{2}{3} still makes sense: it is what each share would be if the total could be split exactly. The mean is a number made from all the values, not one of them.

The values themselves can be fractions too. When they are, add the fractions to get the total and divide by the count, exactly as with whole numbers.

Part 3: working back from the mean

The definition says the mean is the total divided by the count. Turn that around: the total is the mean times the count. If 4 friends have a mean of 6 cards, they must have 4 \times 6 = 24 cards in all, because leveled out they would be 4 towers of 6.

That one fact does two jobs.

  • A missing value. If one value is lost, the total still has to be the mean times the count. Subtract the values you know from that total, and what is left is the missing one.
  • A changed or added value. If one value goes up by some amount, the total goes up by that much, and the extra is shared among all the values. Adding a new value raises the count by one, so find the new total and divide by the new count.

Results

The mean. For a list of n values,

\text{mean} = \frac{\text{total}}{n}

and, turned around,

\text{total} = \text{mean} \times n.

Where the mean sits. The mean is at least the smallest value and at most the largest. It equals them only when every value is the same.

Given equals taken. The amounts by which values sit above the mean add up to the same number as the amounts by which values sit below it.

One value changes. If one value in a list of n values goes up by d, the total goes up by d and the mean goes up by \displaystyle \frac{d}{n}. Going down works the same way.

Why it is true

The mean. Leveling moves cubes from tower to tower without making or losing any, so the total stays the same. When the towers are level there are n equal towers holding the whole total, so each one holds the total divided by n. Multiplying back, n towers of the mean hold the total.

Where the mean sits. If every value were raised to the largest one, the total could only grow or stay the same, and the shared amount would be that largest value. So the real shared amount is at most the largest. Lowering every value to the smallest one shows in the same way that the mean is at least the smallest. If even one value differs from the others, some tower must give and some must take, so the level is strictly between the smallest and the largest.

Given equals taken. Every cube that leaves a tower above the level is placed on a tower below it, and leveling ends only when every gap is filled. So the cubes given and the cubes taken are the same cubes, counted twice.

One value changes. Raising one value by d adds d to the total and leaves the count at n. The new mean is \displaystyle \frac{\text{total} + d}{n} = \frac{\text{total}}{n} + \frac{d}{n}: the old mean plus an equal share of the extra.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, and Examples 5 to 7 with Part 3.

Example 1

Leveling the card stacks

Chiamaka, Ilker and Noelani have 7, 2 and 6 cards. Find the mean, and say who gives and who takes when the stacks are leveled.

Solution

The total is 7 + 2 + 6 = 15 cards. There are 3 friends, so the mean is

15 \div 3 = 5 \text{ cards}.

To reach 5 each, Chiamaka gives 7 - 5 = 2 cards and Noelani gives 6 - 5 = 1 card. Ilker takes 5 - 2 = 3 cards. Check: given 2 + 1 = 3, taken 3, so the cards come out level, as the first figure shows.

Example 2

Cards in a binder

Noelani keeps her cards in a binder. Its 8 pages hold 9, 12, 7, 10, 12, 8, 11 and 11 cards. Find the mean number of cards on a page.

Solution

Add the eight values. Grouping them in pairs that are easy to add uses the commutative and associative properties of addition, which let you add in any order:

\begin{aligned} &(9 + 11) + (12 + 8) + (7 + 11) + (10 + 12) \\ &= 20 + 20 + 18 + 22 = 80. \end{aligned}

There are 8 pages, so the mean is 80 \div 8 = 10 cards on a page. If Noelani moved cards from page to page until every page held the same number, each would hold 10.

Check by leveling. The pages above 10 have extra: 12, 12, 11 and 11 give 2 + 2 + 1 + 1 = 6. The pages below 10 are short: 9, 7 and 8 take 1 + 3 + 2 = 6. The page with 10 neither gives nor takes. Given equals taken, so 10 is right.

Example 3

A share with cookies left over

Chiamaka bakes 5 cookies, Ilker bakes 8 and Noelani bakes 4. They put them all in a cookie jar and share them out evenly. How many does each get?

Solution

The total is 5 + 8 + 4 = 17 cookies for 3 friends. Dividing, 17 \div 3 is 5 with 2 left over, since 3 \times 5 = 15 and 17 - 15 = 2. Each friend gets 5 whole cookies.

The 2 leftover cookies are shared too. Cut each into thirds: that gives 6 thirds, and 6 thirds among 3 friends is 2 thirds each, as the second figure shows. So each friend gets

5\frac{2}{3} \text{ cookies},

which is the mean. Check: \displaystyle 3 \times 5\frac{2}{3} = 15 + \frac{6}{3} = 15 + 2 = 17. Nobody baked \displaystyle 5\frac{2}{3} cookies; the mean is a fair share, not one of the values.

Example 4

Flour for four batches

Ilker makes four batches of cookies and uses \displaystyle 2\frac{1}{4}, \displaystyle 1\frac{3}{4}, \displaystyle 2\frac{1}{2} and 3 cups of flour. What is the mean amount of flour in a batch?

Solution

Add first. The two quarter amounts pair up well: \displaystyle 2\frac{1}{4} + 1\frac{3}{4} = 3\frac{4}{4} = 4. Then \displaystyle 4 + 2\frac{1}{2} + 3 = 9\frac{1}{2} cups in all.

Share \displaystyle 9\frac{1}{2} cups among 4 batches. Split the total into 8 cups and \displaystyle 1\frac{1}{2} cups. The 8 cups give 8 \div 4 = 2 cups each. The \displaystyle 1\frac{1}{2} cups are \displaystyle \frac{3}{2} cups, and \displaystyle \frac{3}{2} \div 4 = \frac{3}{8}, because cutting each half cup into 4 equal parts makes eighths. So the mean is

2 + \frac{3}{8} = 2\frac{3}{8} \text{ cups}.

Check by multiplying back: \displaystyle 4 \times 2\frac{3}{8} = 8 + \frac{12}{8} = 8 + 1\frac{4}{8} = 9\frac{1}{2}. Check by leveling, in eighths: \displaystyle 2\frac{1}{2} is \displaystyle \frac{1}{8} above the mean and 3 is \displaystyle \frac{5}{8} above, so \displaystyle \frac{6}{8} is given; \displaystyle 2\frac{1}{4} is \displaystyle \frac{1}{8} below and \displaystyle 1\frac{3}{4} is \displaystyle \frac{5}{8} below, so \displaystyle \frac{6}{8} is taken. The mean \displaystyle 2\frac{3}{8} is not one of the four amounts, and it sits between the smallest, \displaystyle 1\frac{3}{4}, and the largest, 3.

Example 5

A box that went missing

Ilker sorts his cards into 5 boxes. He remembers that the mean was 24 cards a box. Four boxes hold 21, 30, 18 and 26 cards, but the fifth has gone missing. How many cards were in it?

Solution

A mean of 24 over 5 boxes means the total was 5 \times 24 = 120 cards, because leveled out the boxes would be 5 equal boxes of 24. The four known boxes hold

21 + 30 + 18 + 26 = 95 \text{ cards}.

The fifth box holds the rest of the total: 120 - 95 = 25 cards. (Written as an equation, 95 + m = 120, and undoing the addition gives m = 25, as in One-step equations.)

21 30 18 26 ? 24
Figure 3Ilker’s five boxes have a mean of 24 cards. Four boxes are known; the fifth must bring the total up to 5 \times 24 = 120.

Check by leveling at 24. Above the line, 30, 26 and 25 give 6 + 2 + 1 = 9. Below it, 21 and 18 take 3 + 6 = 9. Given equals taken.

Example 6

A count that was wrong

Noelani writes down how many cookies came out of the cookie jar each day for 4 days: 7, 5, 9 and 3. Then she finds that the 3 was a mistake; that day it was really 11. What was the mean before and after she fixed it?

Solution

Before: the total is 7 + 5 + 9 + 3 = 24, so the mean is 24 \div 4 = 6 cookies a day.

After: the 3 became 11, which is 11 - 3 = 8 more. The total grows by 8, to 32, and the count stays 4. The new mean is 32 \div 4 = 8 cookies a day.

The mean went up by 2, and \displaystyle 2 = \frac{8}{4}: the extra 8 cookies, shared among all 4 days, add 2 to each day’s share. That is the result for one changed value, and it gives the new mean without adding everything again: \displaystyle 6 + \frac{8}{4} = 8.

Example 7

A new binder page

Chiamaka has a binder with 6 pages, and the mean is 9 cards a page. She adds a seventh page. How many cards must go on it for the mean of all 7 pages to be 10? A page has room for at most 18 cards. Could the mean of the 7 pages reach 11?

Solution

The 6 pages hold 6 \times 9 = 54 cards. For 7 pages to have a mean of 10 they must hold 7 \times 10 = 70 cards. The new page must bring 70 - 54 = 16 cards, and 16 fits on a page.

For a mean of 11, the 7 pages would need 7 \times 11 = 77 cards, so the new page would need 77 - 54 = 23. That is more than the 18 a page can hold, so a mean of 11 cannot happen. The most the seventh page can do is 18 cards, which makes the total 72 and the mean \displaystyle \frac{72}{7} = 10\frac{2}{7} cards.

Notice that the answer never needed the six separate page counts. The mean and the count together carry the total, and the total is all the problem uses.

Boundaries

  • The values must be numbers that can be added, all in one unit. Counts of cards, or cups of flour, can be pooled and shared. A list of favorite card colors has no mean, and neither does a list that mixes cups with tablespoons until they are put in the same unit.
  • There must be at least one value. The mean divides by the number of values, and division by 0 has no answer, so an empty list has no mean.
  • Every value counts, including 0. A friend with no cards is still one of the friends sharing. Leaving the 0 out changes the count and gives a different, wrong mean.
  • The mean need not be a possible amount. A mean of \displaystyle 5\frac{2}{3} cards says what the even share would be if the total could be split exactly. It is not a claim that someone has that many.
  • Working backward needs the exact mean and the exact count. The total is mean times count only when both are exact. A mean that was rounded, or a count that is off by one, gives a wrong total and so a wrong missing value.

Where this goes next

The first look at the mean came in Line plots with fractions, where measurements on a line plot were pooled and shared out equally. This lesson gave that equal share its name and its two directions: total divided by count, and count times mean back to the total.

The same number has a second meaning. On a dot plot the mean is the point where the plot would balance, the place where the distances above and the distances below cancel out; that is Mean as balance point. “Given equals taken”, met on the line plot and again in this lesson, is the first glimpse of it. When the same value appears many times, the total can be built from a table of counts instead of a long list, which is Mean from a frequency table and weighted mean.

Exercises

Part 1: finding a fair share

  1. Three stacks of cards have 4, 9 and 5 cards. Find the mean. When the stacks are leveled, which stack gives cards, which take cards, and how many?
  2. Find the mean of 12, 15, 9 and 16.
  3. Over 6 days, Ilker counts the cookies taken from the cookie jar: 5, 2, 7, 4, 6 and 6. Find the mean number of cookies a day.
  4. Find the mean of 30, 45, 25, 40 and 35.
  5. Five friends each have 8 cards. What is the mean? Explain without dividing.
  6. Four classes bring cards to a trading day: 125, 98, 110 and 107 cards. Find the mean number of cards a class.
  7. Find the mean of the stacks 3, 8, 6 and 3. Show that what is given equals what is taken when the stacks are leveled.
  8. Chiamaka has 0 cards, Ilker has 6 and Noelani has 9. Find the mean number of cards per friend. What would you get if you wrongly left Chiamaka out?

Part 2: means that are not whole numbers

  1. Find the mean of 4, 7, 6 and 5.
  2. Three friends put 2, 3 and 5 cookies in a jar and share them evenly. How many does each get? Describe how the leftover cookies are cut.
  3. Five batches of cookies use \displaystyle \frac{1}{4}, \displaystyle \frac{3}{4}, 1, \displaystyle \frac{1}{2} and \displaystyle \frac{1}{2} cup of sugar. Find the mean amount of sugar in a batch.
  4. Noelani measures four cookies across: \displaystyle 2\frac{1}{2}, \displaystyle 2\frac{3}{4}, \displaystyle 3\frac{1}{4} and 3 inches. Find the mean width.
  5. Over four weeks Ilker spends 2.50, 4.00, 3.25 and 1.25 dollars on card packs. Find the mean amount a week.
  6. A list of stacks, each a whole number of cards, has a mean of \displaystyle 4\frac{1}{3} cards. Could there be 4 stacks? Could there be 6? Explain using the total.
  7. Ilker’s counts of cookies over 6 days are 4, 4, 5, 6, 7 and 9. Find the mean. Is it one of the counts? Between which two whole numbers does it lie?

Part 3: missing and changing values

  1. Four stacks of cards have a mean of 9. Three of the stacks have 7, 12 and 10 cards. How many cards are in the fourth?
  2. Noelani opens 12 packs of cards, and the mean is 11 cards a pack. How many cards does she have in all?
  3. The list 5, 9, 7, 3, 6 has a mean of 6. The 3 is changed to 13. Find the new mean two ways: by adding again, and by sharing out the change.
  4. Three stacks have 10, 14 and 12 cards. A fourth stack of 16 is added. What is the new mean? What size of fourth stack would leave the mean the same as before?
  5. Chiamaka’s first 4 batches had a mean of 18 cookies. How many cookies must her fifth batch have for the mean of all 5 batches to be 20?
  6. The list 2, 5, 8 has a mean of 5. Add 3 to every value. What is the new mean, and why?

Reasoning

  1. Ilker says the mean of his stacks 3, 5 and 10 is 9. Without finding the mean, use leveling to explain why he must be wrong. Then find the mean.
  2. Noelani says, “If I take one stack away, the mean always goes down.” Give a list that shows she is wrong, and explain when taking a stack away makes the mean go up.
  3. Explain, using towers of cubes, why the mean times the number of values always gives the total.
  4. Four stacks, each a whole number of cards, have a mean of 5. Give an example where no stack has 5 cards and every stack has an odd number of cards. Could all four stacks have more than 5?
Answers
  1. Mean 6. The total is 4 + 9 + 5 = 18 and 18 \div 3 = 6. The 9 stack gives 3; the 4 stack takes 2 and the 5 stack takes 1, so 3 given and 3 taken.
  2. 13. The total is 12 + 15 + 9 + 16 = 52, and 52 \div 4 = 13.
  3. 5 cookies a day. The total is 5 + 2 + 7 + 4 + 6 + 6 = 30, and 30 \div 6 = 5.
  4. 35. The total is 30 + 45 + 25 + 40 + 35 = 175, and 175 \div 5 = 35.
  5. 8. The stacks are already level, so sharing changes nothing. The total is 5 \times 8 = 40, and 40 \div 5 = 8.
  6. 110 cards. The total is 125 + 98 + 110 + 107 = 440, and 440 \div 4 = 110.
  7. Mean 5. The total is 3 + 8 + 6 + 3 = 20, and 20 \div 4 = 5. The 8 gives 3 and the 6 gives 1, so 4 given; each 3 takes 2, so 4 taken.
  8. 5 cards. The total is 0 + 6 + 9 = 15 for 3 friends, and 15 \div 3 = 5. Leaving Chiamaka out gives \displaystyle 15 \div 2 = 7\frac{1}{2}, which is the share for 2 friends, not 3.
  9. \displaystyle 5\frac{1}{2}. The total is 22, and 22 \div 4 = 5 with 2 left, which is \displaystyle 5\frac{2}{4} = 5\frac{1}{2}.
  10. \displaystyle 3\frac{1}{3} cookies each. The total is 10, and 10 \div 3 is 3 with 1 left. The leftover cookie is cut into 3 equal pieces, one piece for each friend.
  11. \displaystyle \frac{3}{5} cup. The total is \displaystyle \frac{1}{4} + \frac{3}{4} + 1 + \frac{1}{2} + \frac{1}{2} = 1 + 1 + 1 = 3 cups, and \displaystyle 3 \div 5 = \frac{3}{5}.
  12. \displaystyle 2\frac{7}{8} inches. The total is \displaystyle 2\frac{1}{2} + 3\frac{1}{4} + 2\frac{3}{4} + 3 = 5\frac{3}{4} + 5\frac{3}{4} = 11\frac{1}{2}. Then \displaystyle 11\frac{1}{2} = 8 + 3\frac{1}{2}; 8 \div 4 = 2 and \displaystyle 3\frac{1}{2} = \frac{7}{2}, with \displaystyle \frac{7}{2} \div 4 = \frac{7}{8}.
  13. 2.75 dollars. The total is 2.50 + 4.00 + 3.25 + 1.25 = 11.00 dollars, and 11.00 \div 4 = 2.75.
  14. Not 4, but 6 yes. The total is the mean times the count. With 4 stacks it would be \displaystyle 4 \times 4\frac{1}{3} = 17\frac{1}{3}, not a whole number of cards. With 6 stacks it is \displaystyle 6 \times 4\frac{1}{3} = 24 + 2 = 26, which is fine. The count must be a multiple of 3.
  15. \displaystyle 5\frac{5}{6}. The total is 35, and 35 \div 6 is 5 with 5 left, so \displaystyle 5\frac{5}{6}. It is not one of the counts; it lies between 5 and 6.
  16. 7 cards. The total must be 4 \times 9 = 36. The three known stacks hold 7 + 12 + 10 = 29, and 36 - 29 = 7.
  17. 132 cards. The total is the mean times the count, 12 \times 11 = 132.
  18. 8. Adding again: 5 + 9 + 7 + 13 + 6 = 40 and 40 \div 5 = 8. Sharing the change: the value rose by 10, and \displaystyle 6 + \frac{10}{5} = 8.
  19. 13, and a stack of 12 keeps the mean. The old mean is 36 \div 3 = 12. With 16, the total is 52 and 52 \div 4 = 13. A fourth stack of 12 makes the total 48, and 48 \div 4 = 12.
  20. 28 cookies. The first 4 batches total 4 \times 18 = 72. All 5 must total 5 \times 20 = 100, and 100 - 72 = 28.
  21. 8. The new list 5, 8, 11 has total 24, and 24 \div 3 = 8. Each value rose by 3, so the total rose by 3 \times 3 = 9, and sharing 9 among 3 values adds 3 to the mean: 5 + 3 = 8.
  22. The mean is 6. Leveled at 9, the 10 stack could give only 1 card, but the 3 and 5 stacks would need 6 + 4 = 10 cards. Given would not equal taken, so 9 is too high. The total is 18, and 18 \div 3 = 6.
  23. Example: 2, 6, 7 has mean 15 \div 3 = 5. Take away the 2: the mean of 6 and 7 is \displaystyle 13 \div 2 = 6\frac{1}{2}, which is higher. A stack below the mean was taking cards in the leveling; once it is gone, nobody needs those cards and the rest share them, so the mean goes up. Taking away a stack above the mean makes it go down, and taking away a stack equal to the mean leaves it the same.
  24. Level the towers. No cube is made or lost, so the level towers still hold the total. There are as many level towers as there were values, and each is as tall as the mean. So the total is the mean added once for each tower, which is the mean times the number of values.
  25. Example: 1, 3, 7, 9. These are all odd, none is 5, and the total is 20, so the mean is 20 \div 4 = 5. All four above 5 is impossible: each would be at least 6, so the total would be at least 24, but a mean of 5 needs a total of 4 \times 5 = 20.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0