How do I evaluate standard deviation? I have collected responses from 85 people on their ability to undertake certain tasks. 
The responses are on a five point Likert scale:
5 = Very Good,
4 = Good,
3 = Average,
2 = Poor,
1 = Very Poor,
The mean score is 2.8 and the standard deviation is 0.54.
I understand what the mean and standard deviation stand for. 
My question is: how good (or bad) is this standard deviation? 
In other words, are there any guidelines that can assist in the evaluation of standard deviation.
 A: Standard deviations aren't "good" or "bad". They are indicators of how spread out your data is. Sometimes, in ratings scales, we want wide spread because it indicates that our questions/ratings cover the range of the group we are rating. Other times, we want a small sd because we want everyone to be "high".
For example, if you were testing the math skills of students in a calculus course, you could get a very small sd by asking them questions of elementary arithmetic such as $3+2$. But suppose you gave a more serious placement test for calculus (that is, students who passed would go into Calculus I, those who did not would take lower level courses first). You might expect a lower sd (and a higher average) among freshman at MIT than at South Podunk State, given the same test.
So. What is the purpose of your test? Who are in the sample? 
A: Short answer, it's fine and a bit lower than I might have expected from survey data.  But probably your business story is more in the mean or the top-2-box percent.
For discrete scales from social science research, in practice the standard deviation is a direct function of the mean.  In particular, I have found through empirical analysis of many such studies that the actual standard deviation in surveys on 5-point scales is 40%-60% of the maximum possible variation (alas undocumented here).
At the simplest level, consider the extremes, imagine that the mean was 5.0.  The standard deviation must be zero, as the only way to average 5 is for everyone to answer 5.  Conversely, if the mean were 1.0 then the standard error must be 0 as well.  So the standard deviation is precisely defined given the mean.
Now in between there's more grey area.  Imagine that people could answer either 5.0 or 1.0 but nothing in between.  Then the standard deviation is a precise function of the mean:  
stdev = sqrt ( (5-mean)*(mean-1))   
The maximum standard deviation for answers on any bounded scale is half the scale width.  Here that's sqrt((5-3)(3-1)) = sqrt(2*2)=2.
Now of course people can answer values in between.  From metastudies of survey data in our firm, I find that the standard deviation for numeric scales in practice is 40%-60% of the maximum.  Specifically 


*

*40% for 100% point scales,  

*50% for 10-point scales and  

*60% for 5-point scales and  

*100% for binary scales


So for your dataset, I would expect a standard deviation of 60% x 2.0 = 1.2.  You got 0.54, which is about half what i would have expected if the results were self-explicated ratings.  Are the skills ratings results of more complicated batteries of tests that are averages and thus would have a lower variance? 
The real story, though, is probably the ability is so low or so high relative to other tasks.  Report the means or top-2-box percentages between skills and focus your analysis on that.
A: If the data is normally distributed, you could see how population is located.


*

*68% of all people lie within 1 standard deviation of the mean (2.26 - 3.34):





*

*95% of all people lie within 2 standard deviations of the mean (1.72 - 3.88):



It tells you how "spread out" your numbers are.
