Propagation of uncertainty when summation symbol is involved I am unsure how to estimate the propagation of uncertainty when there is a summation symbol involved. 
I have the formula (its used to calculate the sauter mean diameter but I will give a simpler example here):
$$
R = \frac{\sum_{i=1}^{N}n_{i}d_{i}^{3}}{\sum_{i=1}^{N}n_{i}d_{i}^{2}}
$$
For the benefit of clarity, lets say I have taken the weight (to the nearest integer) of 5000 individuals, so $N$ = 5000. 
$d_{i}$ is the weight of each individual and $n_{i}$ is the total number of people with weight $d_{i}$. 
The fractional uncertainty in $d_{i}$ is 3%. I am not sure what the uncertainty is in $R$. I can estimate the uncertainty of $d_{i}^3$ using the advice here:
http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf
But if I just then proceed to carry out the methods when variables are multiplied/ divided, I would surely get a huge uncertainty?
 A: A go at it here:
I'm assume all $n$'s and $d$'s are independent, and that there is no error associated with $n$. 
Let's assume $N=2$.
$$
R = \frac{n_1d_1^3 + n_2d_2^3}{n_1d_1^2 + n_2d_2^2}
$$
When raising a value with an uncertainty to a power without an uncertainty, the following applies:
$$
Z=A^p \\
u_Z = \left | pA^{p-1} \right | u_A
$$
A 3% error on $d$ means that the uncertainty of any $d$ is given as $u_d=0.03d$. It follows from this that the uncertainty on $Z=d^p$  is given by
$$
u_Z = pd^{p-1}0.03d = 0.03pd^p
$$
This can all go back into our original equation
$$
R = \frac{n_1(d_1^3 \pm 0.09d_1^3) + n_2(d_2^3\pm0.09d_2^3)}{n_1(d_1^2 \pm 0.06d_1^2) + n_2(d_2^2\pm0.06d_2^2)}
$$
multiplying in the $n$'s
$$
R = \frac{(n_1d_1^3 \pm 0.09n_1d_1^3) + (n_2d_2^3\pm0.09n_2d_2^3)}{(n_1d_1^2 \pm 0.06n_1d_1^2) + (n_2d_2^2\pm0.06n_2d_2^2)}
$$
From your link to uncertainty propagation, when adding uncertainties, we add them in quadrature
$$
R = \frac{(n_1d_1^3+n_2d_2^3) \pm \sqrt{(0.09n_1d_1^3)^2 + (0.09n_2d_2^3)^2 }}
{(n_1d_1^2+n_2d_2^2) \pm \sqrt{(0.06n_1d_1^2)^2 + (0.06n_2d_2^2)^2 }}
$$
We can look at this and turn it back into sum notation
$$
R=\frac{
\left[ \sum_{i=1}^Nn_id_i^3\right] \pm \left[ 0.09\sqrt{\sum_{i=1}^N(n_id_i^3)^2  } \right]
}
{
\left[ \sum_{i=1}^Nn_id_i^2\right] \pm \left[ 0.06\sqrt{\sum_{i=1}^N(n_id_i^2)^2  } \right]
}
$$
We can simplify this as
$$
R=\frac{A \pm u_A}{B \pm u_B}
$$
Now, $A$ and $B$ are not independent. Therefore, their uncertainties add as the relative errors, not the squares of the relative errors.
$$
R = \frac{A}{B} \pm \frac{A}{B}\left( \frac{u_A}{A} + \frac{u_B}{B} \right)
$$
Hope this helps.
