If the standard deviation of your measurements is in the trillions, then rounding to integers will make little difference – probably no practical difference – since the rounding error is so tiny by comparison to the value. If the standard deviation is $1/2$ then you would be throwing away huge amounts of information and your results will be wrong.
Generally you shouldn't round more than you have to until the last step.
And notice that $(20+\tfrac13)\times 3 = 60 + \left(\tfrac13\times3\right) = \text{exactly } 61,$
but $20.33\times3 = 60.99,$
So $61$ is an exactly answer and $60.99$ is a rounded answer.
If you show $61$ and $60.99$ to a person whose grasp of arithmetic is at a naive level, and as which one is rounded, they'll get it wrong.
Michael Hardy
·
2024-07-17T16:41:19Z (3 months ago)