I know that The more systematic errors > the less accurate The more random errors > the less precise

So lets say we have a stopwatch which has a zero error -1 (every reading less then 1 the actual value)

and an old man is supposed to measure 60s

His readings are 65 ,50,54,62 mean value = 65+50+54+62/4 = 57.75

lets say we remove the systematic error(by adding 1 to each) and the readings become 66 ,51 ,55,63 mean value = 58.75

Although Now mean is more closer to the true value but why is not mean = 60 what am i getting wrong?

  • 2
    $\begingroup$ If you toss a coin 10 times will you always get five heads and five tails? $\endgroup$
    – Farcher
    Sep 26, 2018 at 16:43
  • $\begingroup$ @Farcher No , but why is it that the case? is there a random error while tossing a coin? $\endgroup$
    – Rix Vii
    Sep 26, 2018 at 16:49
  • $\begingroup$ In the case of a coin even though the probability of a head (or tail) is 1/2 but that does not mean that you get exactly half heads and half tails. When taking measurement although the actual value is 60 it does not mean that if you take a number of measurements the average of those measurements will be exactly 60. $\endgroup$
    – Farcher
    Sep 26, 2018 at 17:39
  • $\begingroup$ So what would you call it ? a limitation due to probability ? $\endgroup$
    – Rix Vii
    Sep 27, 2018 at 1:52
  • 1
    $\begingroup$ It is a random error. $\endgroup$
    – Farcher
    Sep 27, 2018 at 6:50

1 Answer 1


What you're getting "wrong" is the fact that your final answer does not have an error associated with it, so you cannot say "my value of $58.75$ is wrong". How do you know? You need to estimate an error associated with it. In this case, the easiest way to assign an error is to find the standard deviation in your measurement values, and then divide by the square root of the number of values (this is a standard result in experimental physics) - i.e. the error in your mean value should be: $$\sigma_\text{mean}=\frac{\sigma}{\sqrt{N}}$$ For this experiment, after correcting for your systematic error (which in this case doesn't actually affect $\sigma$) you get $\sigma_\text{mean}\approx3.5$, and so your final answer based on this experiment is $$\bar{t}=58.8\pm3.5 \text{s}$$ Based on that alone, that is a perfectly reasonable result - the "actual answer" of $60$ s lies within that error bound (which we expect should happen ~60% of the time).

How to "improve" this value? Well if it truly is wrong only because of random errors, then the way to get a more accurate and precise result would be to take more readings (4 readings is not very many!) Unfortunately in practice, systematic errors (errors which are not random) are always present and are in general very difficult to deal with (e.g. the old man might have slow reflexes, which would affect the result etc).

Just to answer the question posed in the title and to summarise my answer, there are in general two sorts of errors:

  • Random errors are random, and are calculated in the way given above. They can be reduced by making more measurements, and this will increase both the precision and (assuming no systematic errors) accuracy of your reading.
  • Systematic errors are all other errors (non-random). They are in general difficult to deal with. You can take as many measurements as you like and get a precise result, but if all of your readings are systematically affected, they will in general give an inaccurate mean.
  • 1
    $\begingroup$ I would add that in his first exemple, if he takes more readings he will eventualy reach a mean of 61s with a very low associated error while in the second exemple he would indead reach a mean of 60s. $\endgroup$
    – Maxter
    Sep 26, 2018 at 17:02
  • $\begingroup$ Yes indeed (though I think you mean 59s instead of 61s). This is what makes systematic errors so infuriating - you can do all the statistical analysis you like, but if you have no appreciation of the physics involved, you will never catch your systematic errors! $\endgroup$
    – Garf
    Sep 26, 2018 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.