Unequal averages
Play around with sets of five numbers and see what you can discover about different types of average...
Age
11 to 14
Challenge level
Problem
Unequal Averages printable sheet
This problem follows on from M, M and M so you may want to look at that first.
Here's an interesting set of five numbers: $$2, 5, 5, 6, 7$$ The mean, mode, median and range are all 5.
Can you find other sets of five positive whole numbers where:
Mean = Median = Mode = Range
Can you find sets of five positive whole numbers that satisfy the following properties?
A. Mode < Median < Mean
B. Mode < Mean < Median
C. Mean < Mode < Median
D. Mean < Median < Mode
E. Median < Mode < Mean
F. Median < Mean < Mode
Not all of these can be satisfied by sets of five numbers! Can you explain why?
Show that some of them can be satisfied with sets of just four numbers.
Show that all of them can be satisfied with sets of six numbers.
If you enjoyed this problem, you may be interested in Wipeout
Student Solutions
The mean, median mode and range have been put below as the number to the left of the set(s).
Students from Kellett School in Hong Kong were able to find some sets:
Sean found that these numbers fit the criteria:
2: 1, 2, 2, 2, 3
5: 2, 5, 5, 6, 7
8: 4, 8, 8, 8, 12
Sebastian found:
4: 2, 4, 4, 4, 6
6: 3, 6, 6, 6, 9
Oliver found that the following works too:
5: 3, 4, 5, 5, 8
Becca and Abi from King's Ely were able to add some more sets:
0: 0, 0, 0, 0, 0
4: 2, 4, 4, 4, 6
5: 2, 5, 5, 6, 7
6: 3, 6, 6, 6, 9
7: 4, 6, 7, 7, 11
8: 3, 8, 8, 10, 11
9: 4, 9, 9, 10, 13
They were also able to deduce that there were no sets possible with mean, median, mode and range all being 1 or all being 3.
Students from Woodcot Primary School in Gosport found almost all the solutions in which the mean, median, mode and range are between 1 and 9:
2: 1, 2, 2, 2, 3
4: 2, 4, 4, 4, 6
5: 3, 4, 5, 5, 8
5: 2, 5, 5, 6, 7
6: 3, 6, 6, 6, 9
7: 4, 6, 7, 7, 11
7: 3, 7, 7, 8, 10
8: 3, 8, 8, 10, 11
8: 4, 8, 8, 8, 12
8: 5, 6, 8, 8, 13
9: 5, 8, 9, 9, 14
George, from St Gabriel's High School in Bury also found:
9: 4, 9, 9, 10, 13
Kirstie, from Strand on the Green Junior School then extended this to find all the solutions up to 13:
10: 4, 10, 10, 12, 14
10: 5, 10, 10, 10, 15
10: 6, 8, 10, 10, 16
11: 4, 11, 11, 14, 15
11: 5, 11, 11, 12, 16
11: 6, 10, 11, 11, 17
11: 7, 8, 11, 11, 18
12: 5, 12, 12, 14, 17
12: 6, 12, 12, 12, 18
12: 7, 10, 12, 12, 19
13: 5, 13, 13, 16, 18
13: 6, 13, 13, 14, 19
13: 7, 12, 13, 13, 20
13: 8, 10, 13, 13, 21
She then considered some larger numbers:
Then I skipped to 100 to see if it would work. I started with a smallest number of 50 which would give 50,100,100,100,150. It also works with 40,100,100,120,140. I then kept going lower until 34 which is the minimum possible for the smallest number. You can also go up until 66 which is the maximum possible.
100: 34, 100, 100, 132, 134 or 35, 100, 100, 130, 135 or ”¦”¦..
or 65, 70, 100, 100, 165 or 66, 68,100,100,166
At this point I noticed a pattern. If you add the minimum and maximum smallest numbers they add up to the number itself.
For 100, minimum = 34, maximum = 66, 34 + 66 = 100.
This is also true for all the other numbers apart from 2, 4 and 6 which only have one solution.
Antonia and Sophie, from Priestlands School, found some instructions to produce sets of five numbers that always have the same mean, median, mode and range:
If the minimum you pick is too large or too small then you get a fifth number that does not fit within your range.
If the minimum you pick is too large, say 7, then the maximum is 15. This means the final number needs to be 2, which is not between 7 and 15, so the range will be 13, which is too large.
If the minimum is too small, say 2, then the maximum is 10. This means the final number needs to be 12, so the range will be 10, which is too large.
Zach was able to explain how Antonia and Sophie should choose their minimum:
Write the five numbers in order as $A$, $B$, $C$, $D$ and $E$.
Suppose the mean is $x$. Since the mean must equal the median, the median must be $x$, which must occupy position $C$.
As the median and mode are equal, the mode is $x$ as well, so at least one of $B$ or $D$ must be $x$.
If $n$ occupies position $A$, then $n+x$ will occupy position $E$, as the range is the same as the mode, so is also $x$.
Since the mean of the numbers is $x$, $A+B+C+D+E=5x$.
This means that the sum of $A$, $E$ and the remaining number that is not set as $x$ must be $3x$.
This means that the missing number must be $3x - n - (n+x) = 2x - 2n$.
This number must lie between $n$ and $n+x$.
If it is equal to either of these, then the set will be bimodal, so will not technically fulfil the criteria.
This tells us that the missing number $2x-2n$ must satisfy these two conditions:
$2x-2n > n$,
$2x-2n < n + x$
If $2x-2n> n$, then $2x > 3n$, and therefore $n < \frac{2x}{3}$.
If $2x-2n < n + x$, then $x < 3n$, and therefore $n > \frac{x}{3}$.
Combining these gives $\frac{x}{3} < n < \frac{2x}{3}$.
This tells us that, if we are going to use Antonia and Sophie's instructions, then we need to choose a minimum value in this range.
If you look back at all the sets listed above, you will see that the minimum number always satisfies this condition.
Joshua and Matthew, also from Priestlands School, and Zach were all able to give good explanations of their methods for all six orders of mean, median and mode, including why two of them are not possible. Take a look at Joshua and Matthew's solution as a Powerpoint or a PDF .
Thomas and Jamie, again from Priestlands School, made a good observation about the order of the mean, median and mode:
The first thing we did was to decide on a mode. If the mode needed to be the highest value then the highest two numbers have to be the same. If the mode had to be the lowest then the lowest two numbers would have to be the same. If the middle value is the mode, then it cannot be completed, because the mode would be the same as the median therefore making them equal rather than bigger or smaller.
Thank you also to Dan and Owen for their contribution.
Well done to everyone who submitted solutions for this problem!
Students from Kellett School in Hong Kong were able to find some sets:
Sean found that these numbers fit the criteria:
2: 1, 2, 2, 2, 3
5: 2, 5, 5, 6, 7
8: 4, 8, 8, 8, 12
Sebastian found:
4: 2, 4, 4, 4, 6
6: 3, 6, 6, 6, 9
Oliver found that the following works too:
5: 3, 4, 5, 5, 8
Becca and Abi from King's Ely were able to add some more sets:
0: 0, 0, 0, 0, 0
4: 2, 4, 4, 4, 6
5: 2, 5, 5, 6, 7
6: 3, 6, 6, 6, 9
7: 4, 6, 7, 7, 11
8: 3, 8, 8, 10, 11
9: 4, 9, 9, 10, 13
They were also able to deduce that there were no sets possible with mean, median, mode and range all being 1 or all being 3.
Students from Woodcot Primary School in Gosport found almost all the solutions in which the mean, median, mode and range are between 1 and 9:
2: 1, 2, 2, 2, 3
4: 2, 4, 4, 4, 6
5: 3, 4, 5, 5, 8
5: 2, 5, 5, 6, 7
6: 3, 6, 6, 6, 9
7: 4, 6, 7, 7, 11
7: 3, 7, 7, 8, 10
8: 3, 8, 8, 10, 11
8: 4, 8, 8, 8, 12
8: 5, 6, 8, 8, 13
9: 5, 8, 9, 9, 14
George, from St Gabriel's High School in Bury also found:
9: 4, 9, 9, 10, 13
Kirstie, from Strand on the Green Junior School then extended this to find all the solutions up to 13:
10: 4, 10, 10, 12, 14
10: 5, 10, 10, 10, 15
10: 6, 8, 10, 10, 16
11: 4, 11, 11, 14, 15
11: 5, 11, 11, 12, 16
11: 6, 10, 11, 11, 17
11: 7, 8, 11, 11, 18
12: 5, 12, 12, 14, 17
12: 6, 12, 12, 12, 18
12: 7, 10, 12, 12, 19
13: 5, 13, 13, 16, 18
13: 6, 13, 13, 14, 19
13: 7, 12, 13, 13, 20
13: 8, 10, 13, 13, 21
She then considered some larger numbers:
Then I skipped to 100 to see if it would work. I started with a smallest number of 50 which would give 50,100,100,100,150. It also works with 40,100,100,120,140. I then kept going lower until 34 which is the minimum possible for the smallest number. You can also go up until 66 which is the maximum possible.
100: 34, 100, 100, 132, 134 or 35, 100, 100, 130, 135 or ”¦”¦..
or 65, 70, 100, 100, 165 or 66, 68,100,100,166
At this point I noticed a pattern. If you add the minimum and maximum smallest numbers they add up to the number itself.
For 100, minimum = 34, maximum = 66, 34 + 66 = 100.
This is also true for all the other numbers apart from 2, 4 and 6 which only have one solution.
Antonia and Sophie, from Priestlands School, found some instructions to produce sets of five numbers that always have the same mean, median, mode and range:
- Choose a number to be your mean, median, mode and range. (8, for example)
- To get the mode place your chosen number twice in your set (8, 8)
- Choose another number (4, for example) to be the minimum in your set (4, 8, 8)
- Add your chosen number to this ($4 + 8 = 12$), so the set is (4, 8, 8, 12). This makes sure the range is correct
- Multiply the mean by 5 to get the total of all five numbers. ($8 \times 5 = 40$)
- Now you use the number that's left between the total you need and the total so far ($40-32=8$).
- If this fifth number is within your range, you have all 5 numbers to get mean, median, mode and range the same name (4, 8, 8, 8, 12)
If the minimum you pick is too large or too small then you get a fifth number that does not fit within your range.
If the minimum you pick is too large, say 7, then the maximum is 15. This means the final number needs to be 2, which is not between 7 and 15, so the range will be 13, which is too large.
If the minimum is too small, say 2, then the maximum is 10. This means the final number needs to be 12, so the range will be 10, which is too large.
Zach was able to explain how Antonia and Sophie should choose their minimum:
Write the five numbers in order as $A$, $B$, $C$, $D$ and $E$.
Suppose the mean is $x$. Since the mean must equal the median, the median must be $x$, which must occupy position $C$.
As the median and mode are equal, the mode is $x$ as well, so at least one of $B$ or $D$ must be $x$.
If $n$ occupies position $A$, then $n+x$ will occupy position $E$, as the range is the same as the mode, so is also $x$.
| $A$ | $B$ | $C$ | $D$ | $E$ |
| $n$ | Possibly $x$ | $x$ | Possibly $x$ | $n+x$ |
Since the mean of the numbers is $x$, $A+B+C+D+E=5x$.
This means that the sum of $A$, $E$ and the remaining number that is not set as $x$ must be $3x$.
This means that the missing number must be $3x - n - (n+x) = 2x - 2n$.
This number must lie between $n$ and $n+x$.
If it is equal to either of these, then the set will be bimodal, so will not technically fulfil the criteria.
This tells us that the missing number $2x-2n$ must satisfy these two conditions:
$2x-2n > n$,
$2x-2n < n + x$
If $2x-2n> n$, then $2x > 3n$, and therefore $n < \frac{2x}{3}$.
If $2x-2n < n + x$, then $x < 3n$, and therefore $n > \frac{x}{3}$.
Combining these gives $\frac{x}{3} < n < \frac{2x}{3}$.
This tells us that, if we are going to use Antonia and Sophie's instructions, then we need to choose a minimum value in this range.
If you look back at all the sets listed above, you will see that the minimum number always satisfies this condition.
Joshua and Matthew, also from Priestlands School, and Zach were all able to give good explanations of their methods for all six orders of mean, median and mode, including why two of them are not possible. Take a look at Joshua and Matthew's solution as a Powerpoint or a PDF .
Thomas and Jamie, again from Priestlands School, made a good observation about the order of the mean, median and mode:
The first thing we did was to decide on a mode. If the mode needed to be the highest value then the highest two numbers have to be the same. If the mode had to be the lowest then the lowest two numbers would have to be the same. If the middle value is the mode, then it cannot be completed, because the mode would be the same as the median therefore making them equal rather than bigger or smaller.
Thank you also to Dan and Owen for their contribution.
Well done to everyone who submitted solutions for this problem!
Teachers' Resources
This problem featured in an NRICH webinar in December 2020.
You may wish to look at Wipeout as a follow-up activity.
The Inquiry Maths website offers a related activity and explains how it can be used as an inquiry prompt in the classroom.
Here is a photo of some of the contributions offered by delegates at the ATM Conference in 2015.
