= Solution
{c}
The algebraic identity $\widehat\beta_0-\beta_0=n^{-1}\sum_i\epsilon_i$ does not require a <normal distribution> for the errors. Applying the <central limit theorem> to their <sample mean> gives
$$
\frac{\sqrt n(\widehat\beta_0-\beta_0)}{\sigma}\xrightarrow d N(0,1).
$$
Thus, for large $n$,
$$
\boxed{\widehat\beta_0\ \text{is approximately }N(\beta_0,\sigma^2/n).}
$$
Its <mean> $\beta_0$ and <variance> $\sigma^2/n$ remain exact, even though its <normal distribution> is now only an approximation. With the error <probability distribution> unspecified, the formula is retained as an estimator; it is not being asserted to remain a <maximum likelihood estimation> formula.
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