- #1
jumi
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Homework Statement
Using 5 different ammeters, you get the following data (all measured in Amps):
[itex]I_{A} = 128 ± 2[/itex]
[itex]I_{B} = 121 ± 1[/itex]
[itex]I_{C} = 114 ± 8[/itex]
[itex]I_{D} = 120 ± 3[/itex]
[itex]I_{E} = 122 ± 4[/itex]
Calculate the mean current and the standard deviation of the mean.
Homework Equations
Standard Deviation of the mean: [itex]σ_{mean} = \frac{σ_{s}}{\sqrt{N}}[/itex]
where [itex]σ_{s}[/itex] is the standard deviation.
Quadrature sum for error propagation: [itex]Total Error = \sqrt{σ_{1}^{2} + σ_{2}^{2} + σ_{3}^{2} + ...}[/itex]
Error propagation formula: [itex]σ_{p} = \sqrt{(\frac{\partial f}{\partial a})^2σ_{a}^2 + (\frac{\partial f}{\partial b})^2σ_{b}^2 + (\frac{\partial f}{\partial c})^2σ_{c}^2 + ...}[/itex], where [itex]p = f(a,b,c,...)[/itex]
The Attempt at a Solution
To get the mean, I added all the data (using the quadrature formula, too) so I could divide by 5.
I got: [itex]\overline{I} = \frac{605 ± 9.6954}{5}[/itex]
To divide that by 5, I used the error propagation formula and got: [itex]\overline{I} = 121 ± 2[/itex]
To get the standard deviation, I would normally get the variance, [itex]σ_{s}^2 = \frac{1}{N - 1} \sum (y_{i} - \overline{y} )^2[/itex], and take the square root.
But subtracting and squaring each data point with the error propagation would require a lot of arithmetic, and that doesn't seem like the right path...
Is there something I'm doing wrong or an easier method to do this? Or do I just have to grit my teeth and do all the arithmetic...?
Thanks in advance.