Percentage uncertainty and percentage error

[david18]

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 can't 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.

 

[_Mayday_]

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 = \frac{1}{124}\times100 = 0,08