# Quick question concerning the sum of random number of random variables given mean and variance and average

$\DeclareMathOperator{\cov}{cov}$The problem is: Let $X_1, \ldots, X_n$ be independent random variables with mean $µ$ and variance $σ^2$. Let $X¯$ be the average of these n random variables. Find the $\cov(X_i, X¯)$

So I used the formula for $\cov$ and the definition of $X¯$ and got zero as my answer but I'm not sure if that's right. Could anyone check my work? Thank you!

$$\text{Cov}(X_i, \bar X)=\text{Cov}\left(X_i, n^{-1}\sum_{j=1}^n X_j\right)$$ $$=n^{-1}\sum_{j=1}^n\text{Cov}\left(X_i, X_j\right)=n^{-1}\sigma^2$$