If you explain something very specific and obscure on the web, then lots of people will end up finding your site. I hope this helps you in your homework or your work assignment. Just remember that statistics are extremely difficult to do properly and there are lots of humorous examples of how they have been inadvertently abused. There is no substitute for applying common sense and critical thinking to avoid doing stooopid things.
Anyway, this is an obscure mathematical description, but one that almost no one seems to understand. I've tried to explain it correctly, but it is subtle and easy to misunderstand. I've included some comments which show the problems with interpretation
If we have sampled a random variable that is a ratio of two separate factors, there are essentially 3 different types of ratios that can occur.
If we wanted to average these numbers, we can look at them as all being in the same units: thousands of characters per 180 seconds (yes, those are the units). The average would simply be the regular arithmetic mean (x1 + x2 + x3)/3 = 1,400 Kchars per 180 seconds or you could convert it to a more meaningful 7.78 Kcharacters/sec.
This method only works if the denominator is not the principal factor under consideration. This is not at all obvious in when measuring throughputs in data communications (see the next case).
This is the common "arithmetic mean" that everyone uses.
If we want the average number of Kchars/sec, would the answer be 8? Technically no. (If you hate mathematics, skip ahead to the intuitive reason below.)
Mathematically, the reason is that the denomoninators were different and you can't add them together (sort of). The best way to "see" it is to replace our numerical samples with the notation of S1, S2, and S3 seconds. Now when you look at the average we just computed, it becomes
...which is what the value of 8 represents.
(Note: S1= 350sec, S2 = 175sec, and S3=
116.666...sec)
Intuitively, the reason is that they are samples with different weights (i.e. the 12.0 sample represents 3 times as much time as the 4.0 sample and so it should carry more weight in the final result). Now here's the tricky part: If we care more about bits than seconds, then, yes, we can do a regular arithmetic average. Let me explain....
In the old days where people would log into an account, download a fixed size file, then hang up, the bit model made sense. People would always transmit the same number of bits regardless of time. Therefore, each of the samples should carry the same weight. This model still fits for people who download huge files such as game demos which can be many tens of megabytes (the time of the session is mostly based on sending a fixed number of bits). However most users nowadays tend to log in for a fixed amount of time and will browse however many web pages they can fit into whatever time period they have. For this model of use, the three samples should not have the same weight.
What this all means is that if the different samples intuitively should not have the same weights, yet the numerators are all the same, they should be considered as a fixed numerator over a varying denominator.
To solve the case where we fix the numerator and vary the denominator, we need to convert the numbers to the same units. An easy way to do this is to simply invert the numbers and take the arithmetic average. This way, we end up with (1/4 + 1/8 +1/12)/3 seconds per Kchars. Since we want the result in the same units that we started out with, it turns out that we can invert the result: 3/(1/4 + 1/8 + 1/12) = 6.5 and not 8.0 Kchars/sec.
It turns out this is the formula for the "harmonic mean" which is used when you are averaging numbers which are ratios of fixed numerators and varying denominators.
First, it's obvious that we can't simply add them together and divide by three. The reason is that the units are different (seconds per 120 Kchars vs seconds per 1,400 Kchars vs seconds per 16,000 Kchars). Next, even if we flip them upside down and try to average the inverted numbers, we still don't have equal denominators. We can't give equal weighting to an 18 second sample and a 1,800 second sample just as we can't give equal weighting to a 120 Kchars sample and a 16,000 Kchars sample. So neither the arithmetic mean nor the harmonic mean will work here.
The correct way to average these samples is to add the numbers together and divide by the total time: (120 + 1,400 + 16,000)/(18 + 180 + 1,800). This effectively combines them into a single sample. WARNING: This method is not valid if the samples are from different distributions.
I don't know of any particular name for this type of average other than the fact that it involved combining all of the samples into one sample, so I call it a "combined average".