I have the following, with once again self-fabricated values, question:
Let C(t) be the number of cats to arrive at a cat palace within $t (\geq 0)$ minutes. Suppose that C(t) has a poisson distribution with mean $\mu t$, where arrival rate is $\mu = 15$ arrivals per minute. $(a)$ What is the mean and standard deviation of the time until the first arrival? $(b)$What is the probability that the $15^{th}$ arrival occurs within 1 minute?
My logic:
$(a)$ Both $\mu$ and $\sigma = \lambda$, so they both equal $15$, I think this is wrong because of the mention of "until the first arrival".
$(b)$ This is a discrete poisson problem:
$\lambda = 15,x = 15$: $\frac{e^{-15}15^{15}}{15!} \approx .1024$
I think this one is also wrong, even though the percentage doesn't seem absurd, it does seem a little low based on the mean being equal.
Any hints or other help are greatly appreciated! If you need further clarification please say so. ${ }{}$
Note that the inter-arrival times of a Poisson RV is an exponential RV with mean $\frac 1\mu$ and variance $\frac 1{\mu^2}$. This is also true of the time of first arrival.
Secondly, if 15th arrival occurs within 1 minute then we know that the number of arrivals within 1 minute is at least 15. Suppose $X(t)$ is the number of arrivals in time $t$. Then we are interested in $\Pr(X(1)\geq 15)$ where $X(1)$ has a Poisson distribution of parameter $\mu$. Now you have to find the following sum: $$ \Pr(X(1)\geq 15)=\frac{e^{-15}15^{15}}{15!}+\frac{e^{-15}15^{16}}{16!}+\frac{e^{-15}15^{17}}{17!}+\dots\approx 0.534 $$ (here.)