Consider a sample $X_1,...,X_{100}$ from a Poisson(3.5) distribution. What is the approximate probability that the total $$T = \sum_{i=1}^{100}X_i$$ exceeds $360$?
I tried normalizing the function but I don't even think that was the correct way to go about this question.
If $X_1, X_2,\ldots,X_n$ are i.i.d. (independent and identically distributed) random variables, with
$$X_i\sim \text{Poisson}(\lambda)\ \ \text{and}\ \ T=\sum_{i=1}^n X_i$$
it can be easily shown (using Moment Generating Functions or Characteristic Functions) that
$$T\sim \text{Poisson}(n\lambda)$$
Therefore, if $n=100$ and $\lambda=3.5$, $P(T>360)=1-F(360)$ in which $F(x)$ is the Poisson c.d.f. (cumulative distribution function) with parameter $n\times\lambda=100\times 3.5=350$ and
$$P(T>360)=1-F(360)=1-\sum_{k=0}^{360} \frac{e^{-350} 350^k}{k!}=0.28524\ \ \ \text{(Poisson)}$$
[note: computed by the software R using 1-ppois(360,350)]
$T$ density distribution can be approximated by a Gaussian distribution, with parameters identical to the exact Poisson distribution, that is $\mu=n\lambda=350$ and $\sigma^2=n\lambda=350$, as the expected value and variance are identical in Poisson. This route leads to:
$$P(T>360)\approx 1-\Phi(\frac{360-350}{\sqrt{350}})=0.29649\ \ \text{(Gaussian approximation)}$$
In this last expression, $\Phi(x)$ is the cumulative distribution of a standard Gaussian distribution. There are some tricks that can be used to improve the Gaussian approximation, but I'm not going to use them. This approximation, for sake of getting probabilities, was more useful in the past when computer resources were scarce, and the exact computations were done manually. Nowadays most computer programs with some statistical capabilities can compute probabilities considering a computer implementation of the exact Poisson cdf. Even Excel can do that.