As per central limit theorem  $(\bar x- E(\bar x))/ \operatorname{Std}(\bar x)$  follows standard normal distribution with mean 0 and standard deviation of 1.

I.e.  $(\bar x- E(\bar x))/ \operatorname{Std}(\bar x)$ follows $N(0,1)$ or equivalently $\bar x$ follows Normal with mean $E(\bar x)$ and variance $\operatorname{var}(\bar x)$.

Note that $E(\bar x) = \theta$ and as $\operatorname{Var}(x) = \theta$ for a Poisson distribution, 
$\operatorname{Var}(\bar x) = \theta/n$ which basically mean $\bar x$ follows normal distribution with mean $\theta$ and variance $\theta/n$.