Variance of the mean of bernoulli RV

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.

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The variables are probably assumed to be independent. Otherwise you can't know the variance - consider the case where they're all equal vs. the case where they're independent. – Yuval Filmus Mar 21 '11 at 5:20
In general you'll need $E(Y_i Y_j)$ for every pair $i$ and $j$ to compute the variance here. (The assumption of independence is $E(Y_i) E(Y_j) = E(Y_i Y_j)$. – Michael Lugo Mar 21 '11 at 5:38
Yes what I refer here is IID. – user8486 Mar 21 '11 at 5:44
IID = Independent and identically distributed. – Yuval Filmus Mar 21 '11 at 6:33