The weight of a certain type of apples has a normal distribution with expectance $\mu = 100g$ and variance $\sigma^2 = 15^2$
If two apples are picked randomly, what is the probability that one of them weighs more than twice as much the other? The answer is $0.0028$ but I'm not sure how to simplify this information so I can work from a standard normal distribution. Any ideas?
Two approaches are provided: first the general one, then the "shortcut" specific to Normal distributions (Gaussians).
General Approach
Presumably the two apples are picked independently with identical distribution, then joint density is just the direct product $$f_{XY}(x,y) = f_X(x) \cdot f_Y(y)~,\qquad f_X(x) = \frac1{ 15\sqrt{2\pi}} e^{\frac{ -(x-100)^2 }{2\cdot (15)^2}}~,~~ f_Y(y) = \frac1{ 15\sqrt{2\pi}} e^{\frac{ -(y-100)^2 }{2\cdot (15)^2}}$$
The density mathematically extends to negative infinity, but there's no such thing as an apple with negative mass. However, with the standard deviation of 15, practically one can ignore the negative part below $X <0 $ and $Y<0$ since the tail is more than six standard deviations away.
The desired probability is $$\mathcal{P}= \int_{x = 0}^{\infty}\int_{y = 0}^{x/2}f_{XY}(x,y) \,\mathrm{d}y\,\mathrm{d}x + \int_{x = 0}^{\infty}\int_{y = 2x}^{\infty}f_{XY}(x,y) \,\mathrm{d}y\,\mathrm{d}x $$ where the two integrals necessarily give exactly the same result since the density is symmetric with respect to the mean.
The evaluation of the integral can be done on any platform of your choice (
R,Matlab,Mathematica,Maple,etc). For example, Wolfram Alpha (which is the online light version of Mathematica) gives the each piece the value of $0.00143456$ after you type in(copy and paste the whole thing above) and click on the equal sign at the right end of the query ("compute"). For the second piece, the code is
Together they make the desired $0.0028$
Specific Solution for Gaussians
Consider a new random variable $W \equiv Y - 2X$, where $X$ and $Y$ are iid Normal with the common mean $\mu_X = \mu_Y = \mu = 100$ and variance $\sigma_X^2 = \sigma_Y^2 = \sigma^2 = 15^2 = 225 $.
The nice properties of Normal distribution include the fact that we already know the distribution of $W$ is also Normal, and that
$$ W = \mu_W + \sigma_W \cdot Z ~, \quad\text{where $Z$ is standard Normal} \\ \mu_W = \mu_Y - 2 \mu_X = -100 \qquad \sigma_W^2 = \sigma_Y^2 + (-2)^2 \sigma_X^2 = 5\sigma^2 = 1225$$
That is, the standard deviation of $W$ is $\sigma_W = \sqrt{5} \sigma = 15\sqrt{5}$.
Accordingly, half of the desired probability will our upper tail here, given by
\begin{align} \Pr\{ Y > 2X \} = \Pr\{ W > 0 \} &= \Pr\Bigl\{ -100 + 15\sqrt{5} Z > 0 \Bigr\} \\ &= \Pr\Bigl\{ Z > \frac{ 0 - (-100) }{ 15\sqrt{5} } \Bigr\} && \\ &= 1 - \Phi\bigl( \frac{ 100 }{ 15\sqrt{5} } \bigr) && \end{align} where $\Phi$ is the CDF of standard Normal. The result is necessarily the same 0.00143456 as seen above. Note that this is of course still a numerical integration and not something you can do "by hand". A look-up table at the end (or front) of a textbook is just compilation of the numerical integration done earlier and with limited precision.
Just for the record, one can also consider $U \equiv Y - \frac{X}2$ that is again Normal, which mean is $100 - \frac{100}2 = 50$ with the standard deviation of $15\sqrt{ 1 + \frac14}$. The desired region is the lower tail
$$\Pr\{ U < 0 \} = \Pr\Bigl\{ 50 + \frac{15}2\sqrt{5} Z > 0 \Bigr\} = \Phi\bigl( \frac{ 0 - 50 }{ \frac{15 \sqrt{5} }2 } \bigr) = \Phi\bigl( \frac{ -100 }{ 15 \sqrt{5} } \bigr) = 1 - \Phi\bigl( \frac{ 100 }{ 15 \sqrt{5} } \bigr)$$
Note that the 2-dim integration approach in the first section is general and applicable to any $X$ and $Y$ that are non-identical nor independent, while this shortcut works only on Normal distributions that are indepedent.