# Prove relationship between average and marginal value

1. Jun 18, 2012

### Bipolarity

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

2. Jun 19, 2012

### chiro

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!).

3. Jun 19, 2012

### mathman

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.