As per Central tendency ( x̅ - E(x̅))/ Std(x̅) follows standard normal distribution with mean 0 and standard deviation of 1

i.e ( x̅ - E(x̅))/ Std(x̅) follows N(0,1) or equivalently x̅ follows Normal with mean E(x̅) and variance var(x̅).

Note that E(x̅) = θ and as Var(x) = θ for a Poisson distribution Var(x̅) = θ/n which basically mean x̅ follows normal distribution with mean θ and variance θ/n.

Hope this clarifies your answer

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 $\theta/n$.