Expert: Josh - 8/11/2003

Question
Hello:

Is the example below a weighted average or a simple average since the quantities at the given prices are more than one box units?

2 Boxes @ $1.50 = $3.00
2 Boxes @ $1.00 = $2.00
2 Boxes @ $1.25 = $2.50

6 boxes total cost $7.50

$7.50 divided by 6 boxes = $1.25/1 box

The average of $1.25 is still the same if one box were purchased and then the average calculated:  $1.50 + $1.00 + $1.25 = $3.75 divided by 3 boxes = $1.25/1 box.


I thank you for your reply.  

Answer
If you do a plot of "number of boxes" versus "unit price", you will see a uniform distribution. Each unit price forms a separate "class" in statistical terms. Strictly speaking, it should be referred to as a frequency histogram (since it is discrete). If you divide the frequency of each class by the total number of occurrences, you will obtain a probability mass function (p.m.f.), with p vs c, as illustrated below.

i.e., CLASS FREQUENCY PROBABILITY
c1=1.50 f1=2 p1=2/6=0.33
c2=1.00 f2=2 p2=2/6=0.33
c3=1.25 f3=2 p3=2/6=0.33

From this, it is clear that regardless of the actual frequency of each class, as long as they are evenly (or uniformly) distributed, the average cost per item will always be the arithmetic mean (A.M.), which is (c1+c2+c3)/3 in this example. Remember that a simple average is a special case of the weighted average. The W.A. is different from the simple average of {c} when the distribution is non-uniform.