Solution to 1992 Problem 48

The general formula for the variance of a random variable y which is a function of n uncorrelated random variables x_1,x_2,...,x_n is
\begin{align*}\sigma_y^2 = \sum_{i = 1}^n \left[\frac{\partial y}{\partial x_i} \right] \sigma_i^2\end{align*}
where \sigma_i^2 is the variance of the random variable x_i and the partial derivative in brackets is evaluated at x_i equal to the mean of x_i for all i. In this case, n = 2, and y(x_1,x_2,...,x_n) is replaced by F(m,a).
\begin{align*}\frac{\partial F}{\partial a} &= \mu_m \\\frac{\partial F}{\partial m} &= \mu_a\end{align*}
where \mu_m and \mu_a are the means of the measurements of m and a, respectively. Thus,
\begin{align*}\sigma_F^2 = \mu_a^2 \sigma_m^2 + \mu_m^2 \sigma_a^2\end{align*}
\begin{align*}\frac{\sigma_F^2}{\mu_m ^2 \mu_a^2} =  \frac{\sigma_m^2}{\mu_m^2} +  \frac{\sigma_a^2}{\mu_a^2}\end{align*}
\mu_F = \mu_m \mu_a because the measurements are uncorrelated. Therefore,
\begin{align*}\frac{\sigma_F^2}{\mu_F ^2} =  \frac{\sigma_m^2}{\mu_m^2} +  \frac{\sigma_a^2}{\mu_a^2}\end{align*}
Therefore, answer (C) is correct.

return to the 1992 problem list

return to homepage

Please send questions or comments to where X = physgre.