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## GR0877 Problem 14

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) (See a wonderful book by John R. Taylor “An introduction to error analysis”, Chapter 7) If one has $N$ separate measurements of a quantity $x$: $x_1 \pm \sigma_1$, $x_2 \pm \sigma_2$, $\dots$, $x_N \pm \sigma_N$, then the best estimate is the weighted average $\displaystyle x_{wav} = \frac{\sum_{i=1}^N w_i x_i}{\sum_{i=1}^N w_i}$, where $w_i = 1/\sigma_i^2$, and the uncertainty in $x_{wav}$ is $\sigma_{wav} = \left( \sum_{i=1}^N w_i \right)^{-1/2}$. Thus, $\sigma_{wav} = \left( \frac{1}{1^2} + \frac{1}{2^2} \right)^{-1/2} = \sqrt{4/5} = 2/\sqrt{5}$.

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1. 10.11.2012 at 02:53

Thank god you know how to do this. So hard to find this online.

2. 16.10.2013 at 04:41

Yeah! Thanks for building this site!

3. 26.03.2014 at 23:03

i didn’t know this but answered in the following way: if both measurements were made with error 1, then the total error is just $1\sqrt(2)$. So here the error must be greater than $1\sqrt(2)$. Also note two measurements is better than one so error must be smaller than smallest error for an individual measurement, so error must be smaller than 1. Then, by elimination, the answer can only be B.