Wikipedia:Reference desk/Archives/Mathematics/2015 June 8

= June 8 =

How I calculate the average of percentages?
How I calculate the average of percentages?

My googlefu is failing, since its giving just very specific examples answered, that arent good enought for learning.

At this moment I am making like this (but its totally wrong problably): The average of 75%, 50% and 80% would be = (75% of (50% of (80% of 100 )))%= 30% 201.79.88.18 (talk) 13:32, 8 June 2015 (UTC)


 * It depends on what sort of average you need. For a simple arithmetic mean (ordinary average), you just add up the percentages and divide by three (if there are three of them), giving sixty-eight and a third percent, but it looks as though you are trying to find the result of combining percentages by successive application (not normally called an average), and this gives your 30%.  If you tell us the context, we can guide you to the correct method.    D b f i r s   14:52, 8 June 2015 (UTC)


 * Your example is not doing any sort of thing we commonly call an average. What you're doing is actually multiplying the percentages. This would make sense if you were repeatedly draining a bucket of water. If you keep 80% of the full bucket and pour the rest out, then keep 50% of that, then keep 75% of that, you'll have 30% of the water you started with. Now, say we had three classes of 100 students. Once class had 50% pass a test, the other class had 80%, and the third class had 75% pass. To get the total percentage of students who pass, it would make sense to average (mean) the percentages, which means we add them up and divide by the total number of numbers: (75% + 50% + 80%)/3 = 68.33%, so 68.33% of the total student population passed the test. To check this answer, you can work it out another way - since each class had 100 students, there were 75, 50, and 80 students who passed in each class. Then there are 75+50+80=205 total students who passed, and 300 total students. 205/300=0.683, which is also 68.3%.
 * So, percentages are just numbers, and we can take the mean of them just like any other set of numbers. But whether that is the appropriate thing to do depends on what question is being asked! So a mean is the right thing to do for pooling percentages over several sub-populations to get a population-level percentage, but multiplying percentages is appropriate for other cases. SemanticMantis (talk) 14:59, 8 June 2015 (UTC)
 * Of course, if each class has different numbers of students, it makes more sense to take the total number who passed, and divide that by the total number of students. E.g. if one class had one student, with a 100% pass rate (i.e. that student passed), and the other had 99 students, with a 0% pass rate (i.e. they all failed), then the combined pass rate would be 1% ((1+0)/(1+99) - which is very different from the unweighted mean of the percentages (50%). MChesterMC (talk) 15:06, 8 June 2015 (UTC)
 * Example 1 similar with what I am trying to do:


 * Map tile X has 4 adjacent tiles.


 * Two of them are desert tiles (up and down adjacent ones), one of them is a florest tile (tile adjacent to left side of x), and the third is a ocean tile (right of x). Desert has 10% of being near a florest tile, ocean has 20% of being near florest and florest has 80% chance of being adjacent with florest tiles.
 * What is the chance of the tile X being a florest tile?
 * Obviously is not just getting the sum of 4 values (10%, 10%, 20% and 80%) and then div by 4. Because if desert had 0% chance of being adjacent to florest, you would have this math [(0+0+20+80)/4=25%], so you would have 25% chance of tile X being florest one, but tile X is adjacent to desert and a desert tile is never adjacent to a florest one.201.79.88.18 (talk) 17:47, 8 June 2015 (UTC)
 * Ah, ok, then we're getting in to conditional probability. So your map looks like this, right?
 * F=Forest, D=Desert, O=ocean. ? aren't specified, and you want to know the discrete probability distribution of $$X \in \{F,D, O \}$$
 * Now, it's a little unclear what the rules are. But I'll imagine that you're looking at a map generated by some sort of cellular automaton. And when you say "Desert has 10% of being near a florest tile", I interpret that as $$\mathbb{P}(X \;\mathrm{is \; Desert}|X \; \mathrm{has \; Forest \; adjacent})=0.1 = 10\% $$ Is that what you're getting at? You'll have to be a little more careful in describing the rules to get an exact answer. You are right that this won't be a simple average. SemanticMantis (talk) 19:39, 8 June 2015 (UTC)
 * The rules are markov chain rules. Each tile type has a chance of being adjacent to other type of tile, this chance (or %) can be from zero to 100 — Preceding unsigned comment added by 201.79.88.18 (talk) 20:28, 8 June 2015 (UTC)
 * Ok, good, we're getting closer I think, but I also think you need to fully specify the markov chain that makes the map before you can compute $$\mathbb{P}(X=Y, \mathrm{for \; each \;} Y \in \{ F,D,O,\}|(X|(X adj O) and (X adj F) and (X adj D) and (X adj D)); $$ which is the formalization for what you're asking. Are you scanning it out line by line? The conditions you give are sufficient to form a 1-dimensional markov chain, but there are several ways to turn that into a 2D array. Does that make sense? This is far more complicated than a simple "average of percentages" :)SemanticMantis (talk) 23:03, 8 June 2015 (UTC)
 * Ok, good, we're getting closer I think, but I also think you need to fully specify the markov chain that makes the map before you can compute $$\mathbb{P}(X=Y, \mathrm{for \; each \;} Y \in \{ F,D,O,\}|(X|(X adj O) and (X adj F) and (X adj D) and (X adj D)); $$ which is the formalization for what you're asking. Are you scanning it out line by line? The conditions you give are sufficient to form a 1-dimensional markov chain, but there are several ways to turn that into a 2D array. Does that make sense? This is far more complicated than a simple "average of percentages" :)SemanticMantis (talk) 23:03, 8 June 2015 (UTC)