on average, a biker inflates their bike tires every 8 days. the interval between each time, namely t1, t2, and so on, is an exponential random variable. Find probability that 40 inflations are required in 1 year.
I know that an exponential distribution is like a poisson distribution but the former is continuouos. Therefore, the modus operandi is pretty the same, and it's straightforward. divide a line into steps, T1 -> T2 -> ... -> Tn, each step is an exponential distribution with 1/8 as parameter k. The expected value of exp(1/8) is equal to 8. and variance is equal to 64.
Now, I have the difficult part. I have to find P(X >= 365). I don't know how to continue.
By watching the solution (because I have the solution), I know that I should've used probability of normal distribution. P(N(0, 1)) and say that this probability distribution is greater than or equal to $$ (365-n*8) / (\sqrt(n*64)) $$
why do I need normal distribution in this exercise? I think my problem is in the logic of the exercise. What do you think about this choice of using normal distribution? why should I use it, and are there other ways to do that? and, why do I have the other part: "$(365-n*8) / (\sqrt(n*64)) $"?
this is the full solution, I've posted it because I think it makes the answer easier to understand, 
EDIT: let's call a generic interval, T. T is an exp(1/8), so by plugging 1/8 in exp formula, and plugging 40 in x, I have $8.4224 * 10^(-4)$. now, I have to multiply it by 8, and I obtain 6.7379 * 10 ^ (-3). I think this is the correct solution.
We know that the expected time between two tyre inflations is $8$ days, or $8/365\approx 0.0219$ years. Therefore the parameter of the exponentially distributed $T_i$ s is $\lambda=365/8=45.625$. The unit of this $\lambda$ is tyre inflations per year. It is known that the number of tyre inflations that are happening before time $t$ is a Poisson process $N_t$ and it is also known that this process has distibution $$\tag{1} \mathbb P(N_t=k)=\frac{(\lambda t)^k}{k!}e^{-\lambda t}\,. $$ The exercise asks for the probability $$\tag{A} \mathbb P(N_t=40)\text{ where }t=1\text{ (year), } $$ or, (there is ambiguity in the exercise) $$\tag{B} \mathbb P(N_t\ge 40)=1-\mathbb P(N_t<40)=1-\sum_{k=0}^{39}P(N_t=k) $$ Using formula (1) can you calculate both ?
I get for (A) a probability of 22.71% and for (B) I get 81.67%.
Why did I use Poisson ?
Because we were asked for the number of tyre inflations per year. The introduction using exponential times $T_i$ is just a test if you know the relationship beteen them and the counting (Poisson) process.
Now to the normal distribution. We know that the expectation and variance of $N_t$ are both equal to $\lambda t$. Then using the standard normal CDF I get the equivalent of (B) as $$ 1-\Phi\Big(\frac{39-\lambda}{\sqrt{\lambda}}\Big)\approx 83.67\%. $$ Close to 81.67% as expected.