Solution to 2008 Problem 14

The weighted average of two independent measurements m_1 \pm \sigma_1 and m_2 \pm \sigma_2 is defined as the value of m that maximizes the log likelihood function
\begin{align*}L\left(m \right)= \log \left\{\frac{1}{\sqrt{2 \pi \sigma_1^2}}\exp \left(- \frac{(m_1 - m)^2}{2 \sigma_1 ^2} \...
or minimizes the function
\begin{align*}\chi^2 \left( m\right) = \frac{\left(m_1 - m\right)^2}{\sigma_1^2} + \frac{\left(m_2 - m\right)^2}{\sigma_2^2}\...
Taking the derivative of \chi^2 with respect to m, setting it equal to 0, and solving for m gives
\begin{align*}m = \frac{m_1/\sigma_1^2 + m_2/\sigma_2^2}{1/\sigma_1^2 + 1/\sigma_2^2}\end{align*}
We use error propagation to find the error \sigma on m:
\begin{align*}\sigma &= \sqrt{\left(\frac{\partial m}{\partial m_1}\right)^2 \sigma_1^2 + \left(\frac{\partial m}{\partia...
We are given that \sigma_1 = 1 \text{ kg} and \sigma_2 = \text{ kg}. Therefore, \sigma = \boxed{\sqrt{4/5} \text{ kg}}. Therefore, answer (B) is correct.

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