There are lots of salmon in a pond and their length (in centimeters) obeys normal distribution $N(70, 5.4^2)$. You and your friend go fishing and decide to continue fishing until both of you catch at least one salmon. What is the probability that your first salmon is $10$ cm longer than your friend's first salmon?
Right answer: $0.10$
I started doing it like this:
$$\begin{align} &E(X) = 70\\ &\sigma = 5.4\\ &Z = \frac{X2 - X1 - E(X)}{D}\\ &P(X2 - X1 \geq 10) = 1 - P(X2 - X1 \leq 10)\\ &= 1 - P(Z \leq (10 - 70)/5.4 ) = 1 - \Phi(-11.11) = 1 - (1 - \Phi(11.11)) = \Phi(11.11) \end{align} $$
Obviously this is wrong because even if I knew the value of $\Phi(11.11)$ it would be something like $0.999\ldots$ and not $0.10$ like it should be.
Any ideas what is the right way to do this?
The normal distribution has the property that the distribution of a sum of independent (not necessarily identically distributed) normal random variables is itself normal with the following properties: $$ \begin{align} \textrm{If}\;&X_i \sim N(\mu_i, \sigma^2_i)\\ \textrm{Then}\;&Y = \sum{a_iX_i} \sim N(\sum a_i\mu_i, \sum a_i^2\sigma^2_i) \end{align} $$ (DeGroot 1989, p.270)
Approaching your problem, what we want is the distribution of the random variable $Y = X_1 - X_2$ where $X_1$ is the length of your fish and $X_2$ is the length of your friend's fish. Given the data, $Y$ should be normally distributed as $Y \sim N\left((70 +(-1\cdot70)), (1^25.4^2 + -1^25.4^2)\right)$ or $Y \sim N(0, 58.32)$.
The probability of having $Y \geq 10$ is then: $$ P(Y \geq 10) = P\left(Z \geq \frac{10 - 0}{\sqrt{58.32}}\right) \approx 1-\Phi(1.309457) \approx 0.09519\;(9.519\%) $$
So either the $10\%$ is an approximation or I've made an error somewhere :)