So, here is the question:
You buy two pieces of pipe from supplier A, and the inner diameter has a normal distribution of N(muA, sigmaA^2) = N(8.02, 0.1^2). You want these two pipes to butt together nicely and not to make ugly turbulent flow.
What is the probability of the diameters of two randomly picked pipes differing by more than 0.3mm?
The crux is that these two points can occur anywhere on the normal distribution, although some places will be more likely, so the area between them will not be fixed. I think it might be something to do with mean values, but I'm absolutely stumped so any help would be greatly appreciated.
Bonus round:
You randomly pick one pipe from the first supplier A, and randomly pick another from a second supplier B, supplier B's inner pipe diameters are described by another normal distribution N(muA, sigmaA^2) = N(7.95, 0.08^2).
What is the probability that the diameter of the two pipes differ by more than 0.3mm?
The answers are apparently 0.03 and 0.04
I'm an engineer-to-be so please be gentle if hardcore maths notation is needed
Good luck and thanks in advance
The basic idea is the following:
Assume you have two independent random variables $X_1 \sim N(\mu_1, \sigma_1^2)$ and $X_2 \sim N(\mu_2, \sigma_2^2)$, then for their difference the following holds:
$$ X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2 )$$
For a proof, see for example the wikipedia article.
Now, for your case, notice that you are looking for the probability $\Pr(|X_1 - X_2| > 0.03)$.
This can be calculated with the corresponding distribution function $F_Z$ as follows (where $Z:= X_1 - X_2$ for ease of notation).
$$\Pr(|Z| > 0.03) = \Pr(Z < -0.03) + \Pr(Z>0.03) = F_Z(-0.03) + 1 - F_Z(0.03)$$
In the first case $Z \sim N(0, 0.02)$, so that the probability is approximately 0.034 and in the second case $Z \sim N(0.07, 0.0164)$ and $\Pr(|Z| > 0.03) \approx 0.038$.