Calculate the Standard Deviation for Multiple Combined Distributions Please see below for the problem.  I've been struggling with this problem for some time and I can't tell if I'm over thinking it...  Obviously the mean reporting error for the engine would remain the same.  But the standard deviation part is confusing me.  How would you calculate the standard deviation when you are selecting 4 or 6 pumps from the distribution data and then evaluating their combined performance?  
Thanks for the help!
"You have the mean and standard deviation data for a population of individual unit pumps' reporting error.  Each engine has either 4 or 6 unit pumps and the total reporting error for the combined performance of all pumps on the engine is being evaluated (i.e four pumps on an engine all of which are over reporting by 10%, the engine reporting error is 10%).  Considering that each pump installed on the engines will be selected from the distribution data provided for an individual pump, what is the overall mean and standard deviation for the entire engine in terms of reporting error for both 4 and 6 pump engines?
 A: Here's my take, assuming I understand the situation correctly.
I think it's important to differentiate between the "real" distribution of error per pump $f_p(E_j;\theta_p)$, which we do not seem to have, and the "measured" statistics of this distribution, of which we seem to have the mean $\mu_p$ and standard deviation $\sigma_p$. From the wording, we seem to have the population statistics for these values; i.e. we can assume they are exactly correct.
Now, let's consider an engine with $k$ pumps. Let's call the error of the engine to be the mean error of its pumps. Let $E_i$ be the error of some pump $i$ and $T$ be the average error of the engine. Then:
$$ T = \frac{1}{k}\sum_i E_i $$
is a random variable, which depends on the random variable $E_i$ for $i$ from $1$ to $k$.
Let's compute the expected value (i.e. mean) of this variable:
$$
\mathbb{E}[T] = \mathbb{E}\left[ \frac{1}{k}\sum_i E_i \right]
= \frac{1}{k}\sum_i \mathbb{E}\left[ E_i \right] = \mu_p
$$
so the expected value is the same.
Let's also compute the expected value of the square which we can use below:
\begin{align}
\mathbb{E}[T^2] 
&= \mathbb{E}\left[ \left(\frac{1}{k}\sum_i E_i\right)^2 \right] \\
&= \frac{1}{k^2}\sum_i \sum_j\mathbb{E}\left[E_iE_j \right] \\
&= \frac{1}{k^2}\left[ k\left(\mathbb{V}[E_i]+\mathbb{E}[E_i]^2\right) + (k^2-k)\mathbb{E}[E_i]\mathbb{E}[E_j] \right]\\
&= \frac{1}{k^2}\left[ k\left(\sigma_p^2+\mu_p^2\right) + (k^2-k)\mu_p^2 \right]\\
&= \frac{\sigma_p^2}{k} +\mu_p^2
\end{align}
Now for the standard deviation. We can compute the variance as
\begin{align}
\mathbb{V}[T^2] &= \mathbb{E}[T^2] - \mathbb{E}[T]^2 = \frac{\sigma_p^2}{k} +\mu_p^2 - \mu_p^2 = \frac{\sigma_p^2}{k}
\end{align}
Thus the standard deviation is:
$$
\text{stddev}(T) = \frac{\sigma_p}{\sqrt{k}}
$$
which makes sense in the sense that the average error should have less variability as $k$ increases (central limit theorem).
In other words, for a $k$-pump engine, the expected error is given by the average error of the individual pumps, whereas the standard deviation is directly proportional to the standard deviation of the pumps, but scaled by the number of pumps involved.
Hopefully that is what you were asking!
