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,
$\begin{align*}\sigma_F^2 = \mu_a^2 \sigma_m^2 + \mu_m^2 \sigma_a^2\end{align*}$
or
$\begin{align*}\frac{\sigma_F^2}{\mu_m ^2 \mu_a^2} = \frac{\sigma_m^2}{\mu_m^2} + \frac{\sigma_a^2}{\mu_a^2}\end{align*}$
$\mu_F = \mu_m \mu_a$ 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.

Solution to 1992 Problem 48