TITLE: Is there something like a normal distribution model for discrete probability?
QUESTION [6 upvotes]: If one wants to find the probability that a continuous random variable will fall within a range of $a \leq X \leq b$, based on a mean value $\mu$, and a deviation of $\sigma$, he would integrate the normal distribution function:
$$\int^b_a \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}dx$$
Since this is for continuous probability, is there an alternative to normal distribution for discrete probability?

Suppose $\mu = 100$, and $\sigma = 50$. For discrete probability, I would try to use bounds close together to achieve a similar, but still not the completely desired outcome. However, the probability is very low:
$$\int^{99.95}_{100.05}p(x)dx = 0.0008$$
The probability seems too low for this to be true, which suggests that a different model for discrete probability should exist.

REPLY [4 votes]: You can construct a discrete distribution called the Skellam distribution (https://en.wikipedia.org/wiki/Skellam_distribution) from the difference of two Poisson distributions. 
If $X \sim Pois(\lambda_1)$ and $Y \sim Pois(\lambda_2)$, then the distribution for $Z = X-Y$ fits the Skellam distribution with mean $\lambda_1-\lambda_2$ and variance $\lambda_1+\lambda_2$. The distribution is approximately normal with the PDF given by $Pr\{Z=n\} = e^{-(\lambda_1+\lambda_2)}\left(\sqrt{\frac{\lambda_1}{\lambda_2}}\right)^n I_n\left(2\sqrt{\lambda_1\lambda_2}\right)$.