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Question

A student attempted to compute the average A of x, y and z by computing the average of x and y, and then computing the average of the result and z. Whenever x < y < z, the student's final result is

Accepted Answer

The true average is A=x+y+z3, and the student computed B=x+y2+z2=x+y+2z4, so B-A = 2z-x-y12 = (z-x)+(z-y)12, which is always positive as z>x and z>y. Thus B is always greater than A, i.e. always greater than A.