How is this random variable distributed? 
Let there be a pipe with inner diameter D and wall-thickness W. They are modulated by independent normal distributed random variables with expected values $\mu_D = 100mm$, $\mu_W=5mm$ and standard deviations $\sigma_D = 0.1mm$, $\sigma_W = 0.05mm$


How is the outer diameter, $D'=D+2W$ distributed?

Since $D$, $W$ are normally distributed, meaning $D,W \sim N(\mu, \sigma^2)$ they have the form...
$$D \sim N(100, 0.1) \Rightarrow F_D(x)= \frac{1}{\sqrt{2\pi\cdot 0.1^2}} \int_0^xe^{-\frac{(x-100)^2}{2\cdot 0.1^2}} dx$$
$$W \sim N(5, 0.05) \Rightarrow F_W(x)= \frac{1}{\sqrt{2\pi\cdot 0.05^2}} \int_{0}^x e^{-\frac{(x-5)^2}{2\cdot 0.05^2}}dx$$
Since they are independent, normally distributed, we know that $D'$ is also normally distributed and in the Ball $B = \{(\tilde d,w): \tilde d+2w = d'\}$. Substitute $u=\tilde d+2w$, $2du =dw$
$$F_{D'} = \int_B f_D(\tilde d)f_W(w)d(\tilde d,w) = 2\int_0^x (\int_{\mathbb{R}} f_D(\tilde d)f_W(u-\tilde d)d\tilde d)du$$
$$2\int_0^x \int_{\mathbb{R}} \frac{1}{2\pi \cdot 0.1\cdot 0.05}\exp(-(\frac{(\tilde d-100)^2}{2\cdot 0.1^2}+\frac{(u-\tilde d-5)^2}{2\cdot 0.05^2})) $$
I simplified that further and got...
$$2\cdot \int_0^x \frac{1}{\sqrt{2\pi(0.1^2+0.05^2)}}\exp(-\frac{(u-100-5)^2}{2(0.1^2+0.05^2)})du$$
Now I know that for normal distributed $X_1, X_2$ the sum $X_1+X_2$ is also normal distributed with $\mu=\mu_1+\mu_2$ and $\sigma =\sqrt{\sigma_1^2+\sigma_2^2}$.
Above I got the same but with an extra $2$ so I don't really know if I did it correctly. Does anybody know?
I guess I accidently computed $D'=2(D+W)$?
 A: $$D \sim \operatorname{Normal}(\mu_D = 100, \sigma_D = 0.1) \\ W \sim \operatorname{Normal}(\mu_W = 5, \sigma_W = 0.05)$$
implies $2W$ is normal with mean $2\mu_W = 10$ and standard deviation $2\sigma_W = 0.1$; thus $D + 2W$ is normal with mean $\mu_D + 2\mu_W = 110$ and variance $$\sigma_D^2 + (2\sigma_W)^2 = 0.02,$$ which corresponds to a standard deviation of $\sqrt{0.02} \approx 0.141421$.
All that integration is unnecessary; you're just increasing the chance of making a calculation error.  Once you know that the sum of independent normal distributions is normal, linearity takes care of the rest.
A: Nothing in this question is really specific to the normal distribution.
By linearity of expectation, $$E(D')=E(D+2W)=E(W)+2E(D)$$  Since $D$ and $2W$ are independent, $$\begin{align}\operatorname{Var}(D')&=\operatorname{Var}(D+2W)\\&=\operatorname{Var}(D)+\operatorname{Var}(2W)\\&=\operatorname{Var}(D)+4\operatorname{Var}(W)\end{align}$$
A: Okay I'm not sure if you've already moved on from this problem, but in case you were wondering your approach was correct, there was just probably some error somewhere.
Define $$X=D+2W$$
Then for $d+2w<x$, we have $w<\frac{x-d}{2}$
The cdf of $X$ is $$\begin{split}F(x)&=\int_{-\infty}^\infty \int_{-\infty}^{\frac{x-d}{2}}f_1(d)f_2(w)\text dw\text dd\end{split}$$
Let $y=d+2w$ so that $\text dy=2\text dw$ (treating $d$ as a constant in the inner integral? I think). We get
$$\begin{split}F(x)&=\frac{1}{2}\int_{-\infty}^\infty\int_{-\infty}^x f_1(d)f_2\left(\frac{y-d}{2}\right)\text dy\text dd\\
&=\frac{1}{2}\int_{-\infty}^x\int_{-\infty}^\infty f_1(d)f_2\left(\frac{y-d}{2}\right)\text dd\text dy\end{split}$$
Taking derivative wrt $x$, we get
$$f(x)=\frac{1}{2}\int_{-\infty}^\infty f_1(d)f_2\left(\frac{x-d}{2}\right)\text dd$$
