Question about unbiased estimate for $P(XLet $X$ be $Normal(\mu,\sigma^2)$, $\sigma$ is known. I need to find an unbiased estimate for $P(X<0)$. My question is, is it enough to find an unbiased estimate for $\mu$ and use that? That is, is $P(X<0|\bar x)$ unbiased for $P(X<0|\mu)$? I'm thinking it is not. If it is not, how do I find an unbiased estimate then?
 A: Your thinking is right, that obtaining an unbiased estimator for $\mu$ is not sufficient, this $\mu \to P(X<0)$ may not be a linear function.
Note that $P(X < 0) = E \left[I(X <0) \right]$, where $I()$ is the indicator function. This representation makes it easier to find an unbiased estimator because for $\frac{1}{n} \sum_{i=1}^{n} f(X_i)$ is an unbiased estimator of $E[f(x)]$.
So, an unbiased estimator of $P(X < 0)$ is $\frac{1}{n}\sum_{i=1}^{n}I(X_i < 0)$ which counts the proportion of observations below 0.
In practice with only one sample, $I(X <0 )$ will either be 1 or 0, and thus will be not a good estimator, although it is still of course unbiased. With only one sample, other estimators will also not be good estimators. JarleTufto's estimator is not valid with one sample since $a$ is not defined.
A: Let $\theta=P(X<0)=\phi(-\mu)$ where $\phi$ is the standard normal cdf.  Assume without loss of generality that $\sigma=1$. 
While dividing the number of observations $X_i<0$ by the sample size $n$ gives you an unbiased estimate of $\theta$ this estimator is likely not efficient because it throws away information.  The MLE on the other hand, $\phi(-\bar X)$, is biased for the reasons given by @Greenparker.
Consider an alternative estimator of the form
$$
\hat\theta=\phi(-a\bar X).
$$
This estimator has expected value
\begin{align}
E\hat\theta
  &=E\phi(-a\bar X) 
\\&= P(Z<-a\bar X) \quad\text{where }Z\sim N(0,1)
\\&=\phi(\frac{-a\mu}{\sqrt{1+\frac{a^2}n}}).
\end{align}
Thus, choosing $a=1/\sqrt{1-\frac1n}$, $\hat\theta$ becomes unbiased for $\theta=\phi(-\mu)$.
Based on the Lehmann-Scheffé theorem, $\hat\theta$ is also UMVUE for $\theta$ since $\bar X$ is a sufficient statistic for $\mu$.
