How to round a discrete random variable? So, I'm currently studying introduction to statistics and I'm on the section about Discrete Random Variables and their distribution tables.
Some questions ask to find the probability within certain ranges (e.g: within one standard deviation from the mean, less than one standard deviation from the mean, etc.)
Since they are discrete variables, we can't have decimals in our $P(x)$ function. But I'm confused on how to round before adding up the probabilities.
For examples, $P(7.22<x<15.99)$, would that be between $7$ and $16$? or $7$ and $15$? And would it make a difference if it was $P(7.22\le x\le 15.99)$? And what about $P(7.99<x<15.22)$? 
Thank you in advance!
 A: In a practical situation, in order to formulate the limits of probabilities such as $P(7.22 < X < 15.99)$ for a random variable $X$ that takes only integer values, you would need a good reason for picking the exact endpoints 7.22 and 15.99. (Why not endpoints
7.23 and 16.00?)
However, once formulated, there can be no question that
$$P(7.22 < X < 15.99) = P(7.22 \le X \le 15.99) 
= \sum_{i=8}^{15} P(X = i).$$
That is, the sum of eight individual probabilities. For example, the
binomial random variable $X \sim Binom(20, 1/2)$ has
positive probability only at integer values. It makes no
difference what non-integer values are included
between the stated endpoints. In particular, for this
binomial random variable, we have $P(7.22 < X < 15.99) = 0.862503,$
computed in R as follows.
 i = 8:15;  sum(dbinom(i, 20, 1/2))
 ## 0.862503

Of course, one could also use the binomial PDF formula to compute
each of the eight terms, and then add them together.
I hope it is now clear that $P(7.22 < X < 16.00)$ and $P(7.22 \le X \le 16.00)$ have different values, unless it happens that $P(X = 16) = 0.$ 
