Binomial distribution approaches to Poisson or Normal? I'm reading two books and they say differently.
In a binomial distribution $X \sim \text{Bin}(n,p)$, if $n \to +\infty$, $X$ approaches to Poisson distribution $\text{Po}(np)$.
The other book says $X$ approaches to normal distribution $\text{N}(np,np(1-p))$.
I'm confused.
 A: Either the two books are sloppy or you're not reading precisely.


*

*If $n\to\infty$ and $p\to0$ while $np$ approaches some positive number $\lambda,$ then the binomial distribution approaches a Poisson distribution with expected value $\lambda.$

*If $n\to\infty$ as $p$ stays fixed, and $X\sim\operatorname{Binomial}(n,p)$ then the distribution of $(X-np)/\sqrt{np(1-p)}$ approaches the standard normal distribution, i.e. the normal distribution with expected value $0$ and standard deviation $1.$
It is sloppy to say something approaches something depending on $n$ as $n\to\infty,$ unless it is precisely defined and not meant literally. Thus the statement that something approaches $\operatorname{Binomial}(np, np(1-p))$ as $n\to\infty$ is not to be taken literally, but rather it means what is stated in the second bullet point above, where the limit, the standard normal distribution, does not depend on $n.$ In the first bullet point above, the statement that something approaches $\operatorname{Poisson}(np)$ can make sense only because $np$ does not depend on $n,$ that is, what is considered is a limit as $n\to\infty$ and $p\to0$ which $np$ remains fixed.
