Mean, median, mode question

[moir]

The median and mode are equal for this set of seven whole numbers: {1, 3, 4, 5,
6, 11, x}. The mean is also one of the six known members of the set. What is the
value of x?

What I already know:
- Of the six given numerals, one of the numerals will be the mean.
- I assume that the x represents a number larger than 11 because the other numbers are in numerical order.

Please help. Thanks!

 

[tkhunny]

I will give you the first one, but how do you KNOW the second? That may be just a coincidence.

Let's think about the median.

(1,3,4)(5,6,11)

Now some conclusions,

If x < 4, then Median = 4 and the Mode is everything (unless x = 1 or x = 3).
If x = 4, then What is the rule for the Median? The Mode is 4.
If 4 < x < 5, then Median = x and the Mode is everything.
If x = 5, then What is the rule for the Median? The Mode is 5.
If x > 5, then Median = 5 and the Mode is everything (unless x = 6 or x = 11).

Where does all that lead us?

 

[lookagain]

I will give you the first one, but how do you KNOW the second? That may be just a coincidence.

Let's think about the median.

(1,3,4)(5,6,11)

Now some conclusions,

If x < 4, then Median = 4 and the \(\displaystyle > > \)Mode is everything (unless x = 1 or x = 3).\(\displaystyle < < \)

If x = 4, then What is the rule for the Median? The Mode is 4.

If 4 < x < 5, then Median = x and the \(\displaystyle > > \)Mode is everything.\(\displaystyle < < \)

If x = 5, then What is the rule for the Median? The Mode is 5.

If x > 5, then Median = 5 and the \(\displaystyle > > \)Mode is everything (unless x = 6 or x = 11).\(\displaystyle < < \)

If there is no repetition of a data value in a set of data,
then there is no mode for that set of data. The mode
would not be "everything."

One of the sources:

http://www.mathgoodies.com/lessons/vol8/mode.html

 

[tkhunny]

If there is no repetition of a data value in a set of data,
then there is no mode for that set of data. The mode
would not be "everything."

I've never been all that comfortable with that hard-line definition. I do find it odd that "bi-modal" is quite acceptable. JeffM's usage of "multi-valent" seems to have ruffled no feathers. Perhaps it would have been more appropriate to say "everything in the set", rather than just "everything". I'm not sure what we gain or lose when we switch from "no mode" to "all members are modes" or back. It sounds the same to me, as long as "all members" does not convey, "feel free to pick one".

If an exam answer key insisted on "no mode", I'd get it wrong on the exam on purpose, just to have the discussion.

 

[soroban]

Hello, moir!

The median and mode are equal for this set of seven whole numbers: {1, 3, 4, 5, 6, 11, x}.
The mean is also one of the six known numbers of the set.
What is the value of x?

What I already know:
- Of the six given numerals, one of the numerals will be the mean.
- I assume that the x represents a number larger than 11
. . because the other numbers are in numerical order. . No!

You can virtually "eyeball" the answer!

The mode is the most populous number.
Since there is only one of each of the given six numbers,
. . the mode must be a duplicate of one of them.


The mean (average) is one of the six numbers.
. . \(\displaystyle \text{Mean} \:=\:\dfrac{1+3+4+5+6+11+x}{7} \:=\:\dfrac{30+x}{7} \)

Since mean is an integer, \(\displaystyle 30+x\) must be divisible by 7.
. . Hence: .\(\displaystyle x \:=\:5,\,12,\,19,\,\hdots\)
And the only one which is a member of the set is: . \(\displaystyle x = 5\)
. . Hence, the mean is: .\(\displaystyle \frac{35}{7} \,=\,5\), which is also a member of the set.


Placing the 7 numbers in order, the median is the middle (4th) number.
. . So we have: .\(\displaystyle 1\quad 3\quad 4\quad 5\quad 5 \quad 6\quad 11 \)
. . . . . . . . . . . . . . . . . . . .\(\displaystyle \uparrow\)


Therefore, \(\displaystyle x\,=\,5.\)
. . And 5 is the mean, median and mode of the set of seven numbers.

 

[tkhunny]

Take 2 billion values

All values have multiplicity 2 - We now have 1 billion modes?

Take 1 billion values.

All are distinct. There is no mode.

Have we accomplished anything, distinguishing the two ONLY because we failed to achieve a multiplicity of 2? I'm struggling to think of a reason why one definition or the other would lead to some meaningful discrepancy.

My views. I welcome others'.

 

[lookagain]

Take 2 billion values

All values have multiplicity 2 - We now have 1 billion modes?

Take 1 billion values.

All are distinct. There is no mode.

Have we accomplished anything, distinguishing the two ONLY because we failed to achieve a multiplicity of 2? I'm struggling to think of a reason why one definition or the other would lead to some meaningful discrepancy.

My views. I welcome others'.

What I responded with is a necessary condition, but not a sufficient condition.
There must be at least one of the values which occurs more than other values.

Current and other examples:


If there are 2 billion distinct values, all with multiplicity 2, then there is no mode.

If there are 2 billion distinct values, with one having a multiplicity of 3, and
all of the rest having a multiplicity of 2, then there is exactly one mode.

Yes, if there are 1 billion distinct values among a total of 1 billion values,
then there is no mode.

-------------------------------------------------


Edit:
-----

Other examples:


A) 1, 1, 2

one mode



B) 2, 2

no mode



C) 1, 2, 2, 3, 4, 4 , 5

bimodal



D) 1, 1, 1, 2, 3, 3, 3, 4, 4, 5, 5, 5

trimodal



E) 1, 1, 2, 2, 3

bimodal



F) 1, 1, 2, 2, 2, 3, 3, 3

bimodal



G) 1, 1, 1, 2, 2, 2, 3, 3, 3

no mode



H) 1, 2, 2, 3, 3, 3, 3, 4, 4, 5

one mode

 

[tkhunny]

Understood. It still seems a little arbitrary.