Understanding the Average: Finding the Middle Number

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Homework Help Overview

The discussion revolves around understanding the concept of average and its relationship to the middle number of a set of values. Participants explore the mathematical definition of average and question its implications, particularly in cases where the average does not correspond to the median of a sequence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to derive the average formula and its connection to the middle number, questioning the logic behind it. Some express confusion about the relationship between the average and the median, while others provide interpretations and alternative formulations.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into the mathematical relationships involved, while others are still seeking clarity on the proof of the average as the middle number. There is no explicit consensus, but productive dialogue is taking place.

Contextual Notes

Participants note that the average may not always represent the middle number in a sequence, leading to further exploration of definitions and mathematical reasoning. The conversation includes references to physical interpretations and alternative formulations of the average.

adjacent
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Homework Statement


Obviously this is not a homework question,but I put this here as this looks like one.
Lets see.
The average of something is [itex]\frac{Sum of all the numbers}{Amount of the numbers.}[/itex]
For example,1,2,3,4,5,6
The average is 3.5 which is also the middle of the sequence.
If the numbers were 1,4,8,9,11
Average is 6.6
Which is not the middle of course.
Anyway,Why does the average give the middle number?

The Attempt at a Solution


Let x and y be integers.
[itex]\frac{x+y}{2}[/itex] =Average
But I don't see the logic behind this.

Also let x and y be integers.and y is the largest number
First y-x to find the range.
So [itex]\frac{y-x}{2}[/itex] gives the middle number between y and x range
So we add the middle number to the x to get the middle number between y and x.
Rearranging,this is same as the average formula.
So is there another way of logicing(1) this?
P.S(1) Couln't think of another word
 
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adjacent said:
So [itex]\frac{y-x}{2}[/itex] gives the middle number between y and x range
No. This gives you the distance between x and the middle number. To get the middle number, you need to add to the distance the starting point. Hence
$$
\frac{y-x}{2} + x = \frac{x+y}{2}
$$
which is indeed the average.
 
DrClaude said:
No. This gives you the distance between x and the middle number. To get the middle number, you need to add to the distance the starting point. Hence
$$
\frac{y-x}{2} + x = \frac{x+y}{2}
$$
which is indeed the average.

Read a little further and you'll see that's what he meant.


A physical interpretation I like is that the average will give you the center of mass. Take a plank of wood and distribute n equal weights amongst it. If you treat the plank of wood as a ruler (distance values from 0-100 say) and then the average of all the values that the weights are located at will be the position where the plank would balance if held up by a pivot.
 
adjacent said:

Homework Statement


Obviously this is not a homework question,but I put this here as this looks like one.
Lets see.
The average of something is [itex]\frac{Sum of all the numbers}{Amount of the numbers.}[/itex]
For example,1,2,3,4,5,6
The average is 3.5 which is also the middle of the sequence.
If the numbers were 1,4,8,9,11
Average is 6.6
Which is not the middle of course.
Anyway,Why does the average give the middle number?

The Attempt at a Solution


Let x and y be integers.
[itex]\frac{x+y}{2}[/itex] =Average
But I don't see the logic behind this.

Also let x and y be integers.and y is the largest number
First y-x to find the range.
So [itex]\frac{y-x}{2}[/itex] gives the middle number between y and x range
As Dr. Claude said, this does not give the middle number.

So we add the middle number to the x to get the middle number between y and x.
This makes no sense. If you already have the middle number to add to the x, adding x will give you something else, not the middle number again.

Rearranging,this is same as the average formula.
So is there another way of logicing(1) this?
P.S(1) Couln't think of another word
I think what you are trying to say is that y- x is the distance between x and y so that [itex]\frac{x- y}{2}[/itex] is half the distance from y to x, not "the middle number". Suppose x and y were the distances from your present location to two cities, X and Y, along a straight road, X being closer than Y (x being less than y). Then [itex]\frac{y- x}{2}[/itex] would be half the distance between them.

To go from your present location to the point halfway between them you must go to X, then half way between X and Y. That is you must go distance x first, then distance (y- x)/2. That gives, as you say,
[tex]x+ \frac{y- x}{2}= \frac{2x}{2}+ \frac{y- x}{2}= \frac{2x+ y- x}{2}= \frac{x+ y}{2}[/tex]
 
ok ok..It's not the middle number,the distance from x to middle number.Sorry.
From the posts of all,I did not see any other way of proving that (x+y)/2 gives the middle number(or median) between them.
How was this formula derived?
 
How about

y - (y-x)/2

:wink:
 
Mentallic said:
How about

y - (y-x)/2

:wink:
That is not a proof.
That is equal to saying the proof of: X*Y gives XY: is YX
:-p
 
I disagree!

adjacent said:
From the posts of all,I did not see any other way of proving that (x+y)/2 gives the middle number(or median) between them.

And the only "proof" given to you so far was that it is equivalent to x+(y-x)/2, hence you're satisfied with the proof that X*Y = XY. Now I'm showing you a whole other world with X*Y = YX :biggrin:

Another idea I have is essentially given in the formula of the mean, but it might lead to some insight that you could have missed.

If we have two numbers x and y, then we're looking for a number z such that x+y = z+z.

So

x+y = z+z
x+y = 2z
(x+y)/2 = z

EDIT: I realize this didn't answer your question. So how can we show that z is in the middle of x and y? Well, all I can think of is to say that y-z = z-x must be satisfied. Given what we found z to be earlier, this equality does indeed hold, hence it's the middle number.
 
Last edited:
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adjacent said:

Homework Statement


Obviously this is not a homework question,but I put this here as this looks like one.
Lets see.
The average of something is [itex]\frac{Sum of all the numbers}{Amount of the numbers.}[/itex]
For example,1,2,3,4,5,6
The average is 3.5 which is also the middle of the sequence.
If the numbers were 1,4,8,9,11
Average is 6.6
Which is not the middle of course.
Anyway,Why does the average give the middle number?

The Attempt at a Solution


Let x and y be integers.
[itex]\frac{x+y}{2}[/itex] =Average
But I don't see the logic behind this.

Also let x and y be integers.and y is the largest number
First y-x to find the range.
So [itex]\frac{y-x}{2}[/itex] gives the middle number between y and x range
So we add the middle number to the x to get the middle number between y and x.
Rearranging,this is same as the average formula.
So is there another way of logicing(1) this?
P.S(1) Couln't think of another word

If ##x < y## (integers or not--it doesn't matter), any number ##z## between ##x## and ##y## can be written in the form
[tex]z = (1-t) x + ty, \;\; 0 \leq t \leq 1[/tex]
When ##t=0## we have ##z=x## and when ##t=1## we have ##z=y##. We can also write this as
[tex]z = x + t(y-x)[/tex]
which means that ##t## is the "fraction of the way from ##x## to ##y##" at the point ##z##. When we put ##t=1/2## we will be at the mid-point of the segment between ##x## and ##y##. Of course, we can re-write the corresponding ##z## as ##z = x + (1/2)(y-x) = (x+y)/2##.
 
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I see.Thank you all
I will think about this further tomorrow. :)))
 

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