Reversing Averages: Extracting Info from an Average

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Discussion Overview

The discussion revolves around the challenge of extracting specific information about the individual inputs that contribute to a given average. Participants explore the implications of averaging a set of whole numbers within a defined range and the limitations of retrieving original data from an average.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether it is possible to determine the exact counts of individual numbers (1's, 2's, etc.) from an average.
  • Several participants assert that there are many combinations of numbers that can yield the same average, indicating a loss of detail in the averaging process.
  • Another participant highlights that the restriction to whole numbers within the range of 1 to 5 may reduce the number of possible combinations compared to a broader set.
  • A later reply introduces the concept of partitions in number theory as a way to express a number as the sum of whole numbers, suggesting a mathematical framework for exploring the problem.

Areas of Agreement / Disagreement

Participants generally agree that it is not possible to uniquely determine the original inputs from an average due to the multiple combinations that can produce the same result. However, there is some discussion about the implications of restricting the inputs to whole numbers, which may lead to fewer possibilities.

Contextual Notes

The discussion does not resolve the mathematical complexities involved in determining the number of combinations that yield a specific average, nor does it clarify the assumptions underlying the participants' claims about the nature of averages and partitions.

fightstacy
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Hello!

I was wondering if anyone had an effective way of extracting information from an average.

I have a list of averages, they're acquired from inputs from 1 - 5, ..and I can see the amount of inputs used to get the average.

An example would be

60 inputs within the range 1 - 5
Average = 2.88

Is there a way to extract how many 1's 2's 3's 4's and 5's were used to get the average from this information?

Thanks in advance!
 
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No. There are many combinations of numbers that produce the same average.
 
Number Nine said:
No. There are many combinations of numbers that produce the same average.

Could I get all possibilities?
 
fightstacy said:
Could I get all possibilities?
No.

When you take the average (mean) of a set of numbers, you lose detail about the numbers.

Suppose you have a very simple set of numbers: {1, 2, 3}. The mean of this set of numbers is 2. This set, {1.1, 2, 2.9} also has a mean of 2, as does {1.01, 2, 2.99}. Any set of three numbers that add up to 6 would have a mean of 2.
 
Mark44 said:
No.

When you take the average (mean) of a set of numbers, you lose detail about the numbers.

Suppose you have a very simple set of numbers: {1, 2, 3}. The mean of this set of numbers is 2. This set, {1.1, 2, 2.9} also has a mean of 2, as does {1.01, 2, 2.99}. Any set of three numbers that add up to 6 would have a mean of 2.

The thing is though, ..that the range 1 - 5 is whole numbers only, no fractions. ..this would surely decrease the amount of possibilities to few, ..am I wrong?
 
fightstacy said:
The thing is though, ..that the range 1 - 5 is whole numbers only, no fractions. ..this would surely decrease the amount of possibilities to few, ..am I wrong?
You are right. You want to know how many ways there are to express N as the sum of M whole numbers. This is the sort of thing that is studied in the theory of partitions, a branch of number theory. See here for more:

http://en.wikipedia.org/wiki/Partition_(number_theory)
 
jbunniii said:
You are right. You want to know how many ways there are to express N as the sum of M whole numbers. This is the sort of thing that is studied in the theory of partitions, a branch of number theory. See here for more:

http://en.wikipedia.org/wiki/Partition_(number_theory)

Thanks for that Jbunniii, looks like a fun read!
 

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