Posted March 4, 2013 Hello!I need help with my physics lab report because i have no idea how to work out the uncertainty. If my experiment gives 3 values for 3 different trials:- 340±14- 350±15- 360±16So that the mean would be 350 but what is the uncertainty of the mean? I did google this and found 3 possible ways of doing this:(360 – 340)/2 = 10 (using the range of the all the values)((360+16) – (340 – 14))/2 = (376 – 326)/2 = 25 (using max and min values)(14 +15 +16)/3 = 15 (using the uncertainties in the values)So what method should I use? Should I quote the answer as 350±10 or 350±25 or 350±15 ? and why?Thanks for your help. Share this post Link to post Share on other sites

Posted March 4, 2013 Hi,the right uncertainty is the second one, so you should quote the answer as 350±25. You can imagine in this way: one of your trial, the highest one, was 360 and the uncertainty was 16, so the maximum possible measured value of the variable in your experiment is 376. You imagine the minimal possible value in your experiment, this means the the 'lowest trial' minus uncertainty, so you end up with 326. You know that uncertainty basically tells you that for example in first trial you could get any value from 326 to 354, because you just add it to or subtract it from the value of variable, right? And by logic you know that all over your experiment you could get any number from 326 to 376, so you just subtract them and divide by 2.Hope it helps PS: I don't know if you're giving just an example but you should do at least 5 trials, especially if the lab is going to be graded on IB criteria) Share this post Link to post Share on other sites

Posted March 4, 2013 I am not an expert at uncertainty. I asked my physics tutor and he told me to take the average of all the values and and take the uncertainty to be the average of the 1'st average and the extremes. Share this post Link to post Share on other sites

Posted March 4, 2013 To find the uncertainty between trials, what you can do is:(Maximum value of a trial - Minimum value of a trial)/2Correct me if i didn't undetstand your question. 1 person likes this Share this post Link to post Share on other sites

Posted March 4, 2013 Hi,the right uncertainty is the second one, so you should quote the answer as 350±25. You can imagine in this way: one of your trial, the highest one, was 360 and the uncertainty was 16, so the maximum possible measured value of the variable in your experiment is 376. You imagine the minimal possible value in your experiment, this means the the 'lowest trial' minus uncertainty, so you end up with 326. You know that uncertainty basically tells you that for example in first trial you could get any value from 326 to 354, because you just add it to or subtract it from the value of variable, right? And by logic you know that all over your experiment you could get any number from 326 to 376, so you just subtract them and divide by 2.Hope it helps PS: I don't know if you're giving just an example but you should do at least 5 trials, especially if the lab is going to be graded on IB criteria)That's incorrect. The largest average you could possibly have is (340+14+350+15+360+16)/3=365.The smallest average you could have is (340-14+350-15+360-16)/3=335. The average is 350, so the uncertainty is +/- 15. When you are dividing numbers like this, just take the accumulated uncertainty and divide it directly. Share this post Link to post Share on other sites