Is there a way to standardize the Poisson distribution? For example, a variable of Normal distribution, $T$, with mean $\mu$ and variance $\sigma^2$ can be standardized into $S$ like this:
$$
S=\frac{T-\mu}{\sigma}\;\Longrightarrow\;F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)
$$
My question is, for the Poisson distribution with probability function
$$
f(k;\lambda)=\Pr(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}.
$$
Is there a way to standardize the $X$ if I define the standard Poisson Distribution as the distribution that $\lambda=1$?
 A: Yes, there is a standard Poisson, the one with parameter $1$. Recall that if $X$  counts the number of "accidents" in a unit time interval, then under suitable conditions $X$ has Poisson distribution with parameter the mean number of accidents per unit time. 
If that parameter is $\lambda$, then the number of accidents in a time interval $t$ is Poisson with parameter $\lambda t$. 
We adjust the interval of time over which we count, until we get a time that gives mean count $1$. Of course $t=\frac{1}{\lambda}$ is that time interval.  
Like in the standardization of the normal, a change of scale is involved. But in the case of the Poisson it is not the random variable that is being scaled.
A: Let X have a Poisson distribution with parameter λ.
Then it can be shown that the random variable Y = (X - λ)/√(λ) will converge to a standard normal distribution as λ goes to infinity.
So if you ever have a possion distribution with a relatively large λ, you can standardize it into Y, which will (approximately) have a standard normal distribution.
