Prove relationship between average and marginal value

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SUMMARY

The discussion focuses on proving that the mean of a new set of numbers, formed by adding a number Y to an existing set of n numbers with an arithmetic mean X, lies between X and Y. The proof involves calculating the new mean as New_Mean = (Sum + Y) / (n + 1) and demonstrating that it is bounded by X and Y. Participants utilize algebraic manipulation to establish the relationship, confirming that the new mean is indeed constrained by the values of X and Y under both conditions where Y is less than or greater than X.

PREREQUISITES
  • Understanding of arithmetic mean and its calculation
  • Basic algebraic manipulation skills
  • Familiarity with inequalities and their properties
  • Knowledge of set theory and number sets
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  • Study the properties of arithmetic means and their applications in statistics
  • Explore algebraic proofs involving inequalities
  • Learn about the implications of adding elements to a data set in statistical analysis
  • Investigate the concept of limits and bounds in mathematical proofs
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Students of statistics, mathematicians, and educators looking to deepen their understanding of mean calculations and proofs involving inequalities.

Bipolarity
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Consider a set of n numbers, and let their arithmetic mean be X.
Suppose a new number Y, is added to the set. Prove that the mean of the new set, with (n+1) members, is between X and Y.

I've been trying to prove this the past few minutes, but I think I'm at a dead end. It's a problem that caught my attention during my reading statistics. Any ideas?

BiP
 
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Bipolarity said:
Consider a set of n numbers, and let their arithmetic mean be X.
Suppose a new number Y, is added to the set. Prove that the mean of the new set, with (n+1) members, is between X and Y.

I've been trying to prove this the past few minutes, but I think I'm at a dead end. It's a problem that caught my attention during my reading statistics. Any ideas?

BiP

Hey Bipolarity.

For first mean we have X = Sum/n. Our new one is given by New_Mean = (Sum+Y)/(n+1) = Sum/(n+1) + Y/(n+1). Now (n+1)/n x Sum/(n+1) = X which implies Xn/(n+1) = Sum(n+1) which implies New_Mean = Xn/(n+1) + Y/(n+1)

So we are comparing X and Xn/(n+1) + Y/(n+1) or X and [1/(n+1)][nX + Y].

Let's assume that our constraints given by the answer are right (i.e the RHS is bound by X and Y inclusive). We will do this to show whether a contradiction occurs either in part or in whole.

I'll let you do this in another post (not a good idea to always post the whole answer!).
 
The expression is (nX+Y)/(n+1)=A
Assume Y < X, then A < (nX+X)/(n+1) = X, A > (nY + Y)/(n+1) = Y.

Similarly for Y > X.
 

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