Question : In a road traffic survey the number of cars passing a marker at a road was counted. The streams of cars in the two directions were a priori modelled as independent Poisson processes of intensities $2 \text{ per minute}$ and $3 \text{ per minute}$, respectively. It was decided to stop the counting once $400$ cars had passed. Let $T_{400}$ be that (random) time point and compute, using appropriate approximations, a time $\alpha$ such that $P(T_{400} \le \alpha) =0.90$
Claimed Answer $85.1$ minutes.
My Attempt : Since $400$ cars passing with a rate of $5$ cars/minute to pass in approximately $\frac{400}5= 80$ min. However this is not correct.
Let $X_i$ be the time between the $i$-th and $(i-1)$-th arrival. Since the sum of two independent Poisson processes in again a poisson process, we know that $X_{i} \sim Exp(1/5)$. Furthermore, we know that the sum of $n$ identical and independent exponential random variables with mean $1/\lambda$ is $Erlang(n,\lambda)$ distributed.
Using this we get that $$\mathbb{P}(T_{400} < a) = \mathbb{P}(\sum_{i=1}^{400} X_{i} < a) = \mathbb{P}(Erlang(400, 1/5) < a) = 0.9.$$ With the R code qgamma(0.9, shape = 400, scale = 1/50) we get that this is the case for $a = 85.16712$.
Edit . My first answer was an exact result. To get an approximate result we can use the central limit theorem. We have that $$ \mathbb{P}\left(\sum_{i=1}^{400} X_{i} < a\right) = \mathbb{P}\left(\frac{\sum_{i=1}^{400} - 400 \cdot \frac{1}{5}}{\sqrt{400} \sqrt{\frac{1}{5^2}}} < \frac{a - 400 \cdot \frac{1}{5}}{\sqrt{400} \sqrt{\frac{1}{5^2}}} \right) = \mathbb{P} (Z < a^{\ast}), $$ where $Z$ is a standard normal random variable and $ a^{\ast} = \frac{a - 400 \cdot \frac{1}{5}}{\sqrt{400} \cdot \frac{1}{5}}$. You can look $\Phi(0.9)$ up in a table and solve $a^{\ast} = \Phi^{-1}(0.9)$ for $a$ to get 85.1 as answer.