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Taking a Weighted Average

```Date: 03/20/2003 at 21:11:54
From: Matt
Subject: Two Different Ways to Take a Weighted Average

I am currently doing a study of truck availability and Use of
Availability for a work project. It involves taking and combining the
following data to get the total number of hours the truck fleet
operates per day:

Truck A: Availability = 80%, Use of Availability = 90%
Truck B: Availability = 90%, Use of Availability = 70%
Truck C: Availability = 95%, Use of Availability = 80%

To determine how many hours each truck operates per day the equation
is (24 * Availability * Use of Availability).

I can't figure out which is the best way to get the average operating
hours for the fleet. Should I find the hours each truck operates and
sum the values or can I find the average Availability and Use of
Availability for the Fleet and get the answer that way?  I tried both
ways and get slightly different numbers.  Is one method incorrect or
is it just a question of significant digits?

Trial #1 - Find individual hours then add together:

Truck A = (24 * 0.8 * 0.9) = 17.28 hours
Truck B = (24 * 0.9 * 0.7) = 15.12 hours
Truck C = (24 * 0.95 * 0.8) = 18.24 hours
Truck Fleet (A + B + C) = (17.28+15.12+18.24) = 50.64

Trial #2 - Find Fleet averages then calculate hours:

Truck Fleet Availability = (0.8*0.9*0.95)/3 = 0.8833
Truck Fleet Use of Avail = (0.9*0.7*0.8)/3 = 0.8

Truck Fleet Hours = (3 * 24 * 0.8833 * 0.8) = 50.88 hours

Thank you!
```

```
Date: 03/21/2003 at 10:59:23
From: Doctor Douglas
Subject: Re: Two Different Ways to Take a Weighted Average

Hi, Matt,

Thanks for submitting your question to the Math Forum.

How to properly take averages in different situations usually comes
down to what question is being asked, or what quantity you want to
estimate. Sometimes this is a tricky business to get right.

Because you asked for "the total number of hours the truck fleet
operates per day," you can simply add up the individual trucks, as you
did in method 1.  This is certainly a "weighted average," because each
truck's use is being weighted by its availability. I think method 1 is
correct.

Now, what is method 2? The average availability is 0.8833, and the
average use (irrespective of availability) is 0.8. You could multiply
these two together, and the answer would provide an estimate of hours
based on an *imaginary* truck whose availability is the average
availability and whose use is the average use. This is probably a more
complicated statistic than you need. It doesn't take into account
correlations between the availability and the use, which could be
important (in fact they probably are - see the example below). And
method 1 - you directly compute the total number of hours (by
computing the hours for each truck) in method 1, not just estimate a
fleet average based on the three imaginary trucks.

To see how the "correlations" I mentioned could be significant,
imagine you have the following three trucks:

availability       use
98%            10%
50%             0%
2%           100%

Method 1 gives the total number of hours for all three trucks as
24(.98*10 + .50*0 +  0.02*1) = 2.832 hours.  But method 2 gives the
following:  24*(.98 + .5 + .02)*(.1 + 0 + 1)/3 = 13.2 hours. How can
this be right? None of the trucks manages more than 24*0.98*0.10 =
2.352 hours individually. The answer lies in how the use and
availability "correlate." The average availability is 50%, and the
average use is a shade above one-third. The use average has been
"skewed" by the high use number of the third truck, whose availability
is only 2% in the first place.  If there had been a fourth truck with
availability 1% and use 100%, this problem would have been even more
dramatic.

The bottom line is that it is best to use the data when available, and
to be careful when doing too much manipulation of "average" values.

I hope this helps.  Please write back if you have more questions

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Statistics
Middle School Statistics

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