Variance of the mean of bernoulli RV

Question

I have the following problem :
Let $X_1,...,X_n$~N($\theta$,1). Define $$Y_i=\begin{cases} 1& X_i> 0\\ 0& X_i \leq 0. \end{cases}$$ Let $\psi=\mathbb{P}(Y_1=1).$ Define $\widetilde{\psi}=\frac{1}{n}\sum_i Y_i.$ Find the standard deviation of $\widetilde{\psi}$
Clearly $\psi=p$ and $Y_i$ are Bernoulli ($p$) then I have $$\mathbb{V}(\frac{1}{n}\sum_i Y_i)=\frac{1}{n^2}\mathbb{V}(\sum_i Y_i)$$ but there is no statement about independence, so it cannot be done the argument that $$\mathbb{V}(\frac{1}{n}\sum_i Y_i)=\frac{1}{n^2}\sum_i \mathbb{V}(Y_i)$$ or the argument that since $Y_i$ are bernoulli of the same variables and independent then they are a binomial RV. How I find the variance here. I really appreciate any help.