## 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.