The problem solving process is as follows:
Remember that the variance is s and the expected value is e.
The first question: EA = (1-3-4+4+2-1+2)/10 = 0eb = (4-3-1+2).
Second question: SA = (1-0) * (1-0)+...+(2-0) * (2-0) =1+9+16+1.
Because s b is smaller, the deviation of B clock will be smaller under the same average error!
The price is the same, of course, choose B clock!
As for the situation you discussed, I don't think you have noticed the characteristics of variance and mean. They are suitable for large sampling values. If the sampling value is small, it loses its meaning, so the argument of expected value and variance is useless. Good luck!