Question : " Customers arrive at a shop such that the number of arrivals in any interval of duration d hours follows a Poisson distribution with mean 8d. The third customer on a particular day arrives T hours after the shop opens.
Here is what we are asked for to find"
I having trouble understanding the concept of the question and what we are trying to find. To my understanding for part i) P(T>t)=P(X=0)+P(X=1)+P(X=2) but I still don't really understand it in detail.
On the other hand for part ii) I understand that I have the cdf and I should differentiate it to get the pdf but when I do so and check my answer I get a negative. What the answer said is -F'(x)= d/dx ( part i).
And finally, for part iii), I can work through both but it says "deduce" and I can't figure out how we are supposed to deduce it.
Any help will be of great appreciation. Thanks!
For part ii, note that $F(t) = P(T<t) = 1 - P(T>t)$ so when you differentiate $P(T>t)$ you get minus what you want.
For part i, we are finding the probability that the 3rd customer arrives after time $t$. This is equivalent to asking that at time $t$, either $0$, $1$, or $2$ customers have arrived.
The probability of this is $e^{-8t}+e^{-8t}(8t)+e^{-8t}(8t)^2/2$. You can see this works out to be right.
Part iii, idea is you can identify the distribution of $T$ from its PDF and then give the mode and mean. For mode I'd probably just maximise the PDF. For mean you can either work it out or match your distribution with a well known one. Probably just work it out.