Can I use z-scores in right skewed distributions? I have a right-skewed distribution that represents that number of "likes" on a certain car category for a number of users. What I am trying to do is a sort of classification based on z-scores. For example, if a user has made a lot of likes in favor of the given category (more than 1 standard deviation above the mean) then the user is considered engaged. Else, if the user lies 1 standard deviation below the mean then the user is not engaged at all. How can I apply this logic on my right-skewed data?
 A: There are two quite distinct questions muddled together here. 
Given $z =$ (value $-$ mean) / standard deviation (SD), your choices of $z > 1$ to mean "engaged" and $z < -1$ to mean "not engaged" are at best practical choices based on some context you don't give. It is difficult for anyone not familiar with your data and situation to comment, except that statistical experience might lead to warnings that your thresholds are arbitrary and are all too likely to separate people with minutely different values. 
Whether the underlying distribution is symmetric or skewed is another question. If the distribution is approximately symmetric, then the fractions classified as "engaged" or "not engaged" will be about the same and can be predicted quantitatively if (and only if) you are willing to consider guesses about what the underlying distribution is (e.g. approximately symmetric binomial). If the distribution is skewed, that almost always will not hold, and it is even possible that no values lie below $z < -1$. Either way, the proof of the classification is in whether it helps some analysis somehow and you can always count how many you have in either class. 
If this kind of data were mine to analyse, I would always look directly at the number of "likes" (presumably ranging from 0 to some maximum). It is not obvious to me that you need classify at all, as people are already classifying themselves; nor is it obvious that mean and SD offer the best descriptive framework. 
A: I would advise against using z-scores for markedly skewed data.  For a skewed distribution, both the mean and the standard deviation are affected by the skew in a way that make the z-score results less representative of what you're trying to convey.
I think a more direct way of indicating what you want is to use percentiles.  Percentiles are in a sense robust against the skew in the distribution.  You could use 50th percentile (the median), and say, for example, those above the median are "more engaged".  Of course, you could also use the 75th percentile, or 90th percentile, as makes sense for your application.
As an example of this application, consider a classic skewed distribution, income, in this case the distribution of U.S. income from 2010.  Note that the figure indicates the 50th (median), 75th, and 90th percentile of income.
As others have mentioned, the practical interpretation of these percentiles require some caution.  You can confidently say that those with an income greater than the median have a higher income than the "average household"†.  But practically speaking --- and living in the U.S. ---, I wouldn't say that those with greater than the median income are "high income"‡.  I also wouldn't say that those with greater than 90th percentile income are "high income".  It's difficult to assess, of course, because the U.S. is highly heterogeneous in this respect, so that $100,000 a year in New York City or San Francisco may not get you a place to live, whereas in some places in the country that's a really comfortable income.

†  This is typical language in this application.
‡  I may have a somewhat skewed perspective since I live in New Jersey where the median income is about $80,000.
