Poisson Thinning and Pareto Distribution was hoping to get some advice on this one. It is a two part question.
The number of days(D) in any given month has a Poisson Distrubtion with $\lambda$ =10 days that rain per month. Let $X_1,X_2$... be independent amounts of rain(mm) that are independent of D and have a Pareto distribution with parameters $\alpha = 5$ $\delta = 30$. Also, let Z= $\sum_{i=1}^D$ $X_i$ be the total rainfall for any given month. 
a, Calculate the mean and standard deviation of Z. 
b, Find the probability that at least one day in a month has 80mm or more rain. (Question suggests looking at Poisson Thinning)
To calculate the mean of Z, I multiplied the mean of the Poisson $\lambda$=10 by the mean of the Pareto, $\delta / \alpha - 1$ = 7.5, to give 16*7.5=120.
For the variance, the same applies, Var Poisson = 10 & Var Pareto = $\frac {\alpha \delta^2}{(\alpha-1)^2(\alpha-2)}$ = 93.75, Var Z= 937.5 and $\sigma$= $\sqrt{937.5}$ =30.6186
My question is with part b. To find the probability that any given day has 80mm or more rain, would I use the cdf of the Pareto? My thinking is that P(X$\gt$80mm) = 1- P(X$\lt$80mm) using this I got the answer, P(X$\gt$80)= 1-(1-($\frac {\delta}{\delta+y})^\alpha$ = 0.0015(seems too low but i'm not sure) Regardless of whether or not this is the answer to the probability, i'm sure there is a step I am yet to complete. This probability is calculated given that it's a rainy day, whereas the question asks, what is the probability that at least one day will have that much rain. Any advice would be greatly appreciated. 
 A: Your calculation of the variance is incorrect, because $$\operatorname{Var}[Z] \ne \operatorname{Var}[D]\operatorname{Var}[X_i].$$  Instead, you need to use the law of total variance: $$\operatorname{Var}[Z] = \operatorname{E}[\operatorname{Var}[Z \mid D]] + \operatorname{Var}[\operatorname{E}[Z \mid D]].$$  In your particular case, $D$ is a counting distribution, with $Z = X_1 + \cdots + X_D$ and the $X_i$ are IID, so the conditional variance of $Z$ given $D$ is simply $$\operatorname{Var}[Z \mid D] \overset{\text{iid}}{=} D \operatorname{Var}[X_i].$$  This is because the variance of a sum of independent random variables is equal to the sum of their variances; and because the variables are identically distributed, each has the same variance; and there are $D$ such $X_i$.  Consequently, $$\operatorname{E}[\operatorname{Var}[Z \mid D]] = \operatorname{E}[D \operatorname{Var}[X_i]] = \operatorname{Var}[X_i]\operatorname{E}[D].$$  Note this expectation is taken with respect to the distribution of $D$, as the variance of $X_i$ is a fixed constant.  Next, we have $${E}[Z \mid D] \overset{\text{id}}{=} D \operatorname{E}[X_i],$$ by linearity of expectation and the fact that the $X_i$ are identically distributed.  Then $$\operatorname{Var}[\operatorname{E}[Z \mid D]] = \operatorname{Var}[\operatorname{E}[X_i] D] = \operatorname{E}[X_i]^2 \operatorname{Var}[D].$$  So putting all this together, $$\boxed{\operatorname{Var}[Z] = \operatorname{Var}[X_i] \operatorname{E}[D] + \operatorname{E}[X_i]^2 \operatorname{Var}[D].}$$  This is the proper formula to use.

As for the second part of your question, the idea is to think of the Poisson events (a rainy day) as belonging to one of two classes:  either on that rainy day, the total rainfall was less than $80$ mm, or the total rainfall was at least $80$ mm.  Then because of Poisson thinning, each of these two classes of events is itself a Poisson process with a Poisson distribution, in which the intensity parameter is some function of the proportion of the respective class of event relative to the overall intensity of occurrence.
For example, if I conduct an experiment in a neutrino detector, in which I model the number of neutrinos passing through the detector per second as following some Poisson distribution with intensity $\lambda$, but I know that on average, only $1\%$ of such neutrinos actually interact in the detector and produce an observable reaction, then the rate of observed events is modeled by a Poisson distribution with intensity $0.01\lambda$.
Therefore, in your case, if you compute the probability that the amount of rain is at least $80$ mm on a given rainy day, you would use this as a way to model the random number of rainy days with rainfall at least $80$ mm in a month.  Then use the Poisson distribution to compute the probability that no such events occur, and then take the complementary probability--which is the probability that at least one such day occurs.
