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Percentage uncertainty and percentage error 
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#1
Feb1208, 12:30 PM

P: 50

You record the following data Diameter/mm: 11.67, 12.67, 13.9, 12.67, 12.67, 12.66 You take the average of these values to be the diameter, but what is the percentage error in this value?
I tried working this problem out but I cant use the normal formula as it is spread over many values and there is no accepted value. I got some numbers; 17.55%, 1.5% but theyr all wrong. W = T(x+y)/x Where W is the weight of the stand and T is the Newton meter reading. You measure and record the values of x, y (with a ruler) and T. The following data is recorded: x = 12.8 cm , y = 30.2 cm, T= 2.2 N What is the percentage uncertainty in y? This one wants the percentage uncertainty which I'm unsure of how to find. I did try 30.25(max value)  30.15(lowest value) / 2 = 0.05 and find it as a percentage but it turned out to be wrong... Any help with these problems would be much appreciated. 


#2
Feb1208, 12:56 PM

P: 816

Percentage Uncertainty
Uncertainties may be quoted as a percentage rather than absolute values. An uncertainty of 124 (+ or ) 1 means 1 in 124 ie. Percentage Uncertainty = [tex]\frac{1}{124}\times100 = 0,08[/tex] 


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