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Prove relationship between average and marginal value

  1. Jun 18, 2012 #1
    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?

  2. jcsd
  3. Jun 19, 2012 #2


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    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!).
  4. Jun 19, 2012 #3


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    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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