The standard deviation of a set of data is useful because any particular data point is likely to be within one or two standard deviations of the mean. For example, if your average is 17s and standard deviation is 2s, then most of your times will be between 15s and 19s. If your standard deviation is 1s with the same average, then most of your times will be between 16s and 18s.
Start by saying that "deviation" is how far any particular time is from your average. Then in (a mouthful of) words, standard deviation is the square root of the average of the square of the deviation. To calculate it, do this:
- Calculate the average
- For each data point, subtract the average from it, then square the answer
- Take the average of the answers you got in the previous step
- Take the square root of that
So, for example, if you do a trial of 5 that is
60
63
52
88
59
then the average is (60+63+52+88+59)/5 = 64.4
subtracting the average from each time, the list becomes
-4.4
-1.4
-12.4
23.6
-15.4
Squaring each of those gives
19.36
1.96
153.76
556.96
237.16
The average of that list is 193.84, and the square root of that is roughly 14. So the standard deviation of that list of five times is about 14.
Five data points is a very short list, though, so most of the time you'd calculate a standard deviation using more data points than that. You'll find that if you track your standard deviation over the course of a 100-solve trial, at first it will vary up and down, but eventually it will settle down to a certain value and pretty much stay there. It shouldn't drift as you do more and more solves (unless you get tired and start messing up more!).
Standard deviation has the same units as the original data, so if you are timing in seconds, standard deviation also is in seconds.
In many data sets, about 70% of the data points are within one standard deviation of the mean, and 95% are within two. We can use this to calculate the chances of certain events happening.
Someone who averages 13s with a standard deviation of 1s might want to know the chances that they'll get a sub-10s solve in competition. This is 3 standard deviations below their mean, and a "3-sigma event" like this has a chance of about 1/1000. If they go to a competition and do 15 speed solves, they have about a 1.5% chance of getting a sub-10, assuming they solve the same way in competition as they do in practice.
However, it's important to note that standard deviation does not tell you everything about your solves. For example, in the previous paragraph I used the numbers that correspond to a bell curve, but if your times do not fall on a bell curve, the calculations could be wrong. An easy way for this to happen is occasionally to have a very bad time. A few bad times will increase the standard deviation a lot (depending on how bad they are, and how large the sample size is). If my original example had replaced the 88s solve with a 58s solve, the standard deviation would fall from 14 to 4.
You can see that this would affect any calculations with the standard deviation severely. Getting the bad time would actually increase your calculated odds of getting a good time! (It doesn't actually increase your chances of getting a good time. It's an error in the calculation.) Another useful figure to calculate, in addition to the standard deviation, would be the skew. Most cubers' solve times probably have positive skew (especially novices), and accounting for this could allow better calculations. In some cases you might also want to know the kurtosis or higher order moments of the distribution.
A useful thing to do with standard deviation is to divide the deviation of one particular data point by the standard deviation, to get a "z-score". This is useful to tell you how unusual an event is. For example, if you see a 6'7" tall man at the airport, and then you have your best solve time ever on a lucky scramble, you might want to know which event is more unusual. You could calculate the z-score of the event of a man being 6'7" tall and the z-score of getting your fast time using the averages and standard deviations of those completely different distributions. Whichever had the higher z-score would be more unusual. This doesn't tell you which was more unlikely to happen. If you saw 500 people at the airport and did 50 solves, then it makes sense that you would see a more unusually tall person than get an unusually fast solve, but the z-score does still tell you how unlikely one particular event is.
Finally, as the previous poster pointed out, fast cubers will have a smaller standard deviation than slow cubers. Mine is roughly 10% of my solve time (I solve in about 40s and have a standard deviation of about 4.)
