Percentage uncertainty and percentage error

  1. Feb 12, 2008 #1
    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. jcsd
  3. Feb 12, 2008 #2
    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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