You are travelling from $A$ to $C$. In order to reach your destination, you have to take two buses and make a transfer halfway at point $B$. The time at which bus 1 is expected to reach $B$ is normal distributed $T_1 \sim \mathcal{N}(\text{12:00}, \text{$5^2$ min})$, and bus 2 is expected to leave at $T_2 \sim \mathcal{N}(\text{12:02}, \text{$3^2$ min})$. What is the probability of making the connection?
I know the answer can be found by $P(T_1 - T_2 < 0)$ and do the $Z$ normalization with $Z = {T1 - T2 - (\text{12:00} - \text{12:02}) \over \sqrt{5^2 + 3^2}}$, but how does one prove this?
This is what I was looking for
I found some help through these questions
https://stats.stackexchange.com/questions/383562/difference-between-two-random-variables
Convolution of two gaussian functions
From the first question we can derive (I'm pretty sure I did the second step wrong)
$$\begin{align} P(T_1 < T_2) &= \int_{-\infty}^\infty\int_{t_1}^\infty p_{T_1}(t_1)p_{T_2}(t_2)\,\mathrm dt_2\mathrm dt_1 \\ &= \color{red}{\int_{-\infty}^\infty\int_0^\infty p_{T_1}(t_1)p_{T_2}(t_2)\,\mathrm d(t_1-t_2)\mathrm dt_1} \\ &= \int_{-\infty}^\infty\int_{-\infty}^0 p_{T_1}(t_1)p_{-T_2}(-t_2)\mathrm dt_1\mathrm d(t_1-t_2) \\ &= \int_{-\infty}^0\int_{-\infty}^\infty p_{T_1}(t_1)p_{-T_2}(t_1-t_2-t_1)\mathrm d t_1\mathrm d(t_1-t_2) \\ &= \int_{-\infty}^0 p_{T_1}\ast p_{-T_2}(t_1-t_2)\mathrm d(t_1-t_2) \\ &= \int_{-\infty}^0 p_{T_1-T_2}(t_1-t_2)\mathrm d(t_1-t_2) \\ P(Z < 0) &= \int_{-\infty}^0 p_Z(z) \mathrm dz \end{align}$$
And from the second question we can derive that the convolution between two normal distributions is
$$ X - Y = \mathcal N(\mu_1, \sigma_1) \ast \mathcal N(\mu_2, \sigma_2) = \mathcal N(\mu_1 - \mu_2, \sqrt{\sigma_1^2 + \sigma_2^2}) = Z $$