From normal dist. with unknown variance, how to approximate $P(\bar X>c)$? From normal dist. with known mean $\mu$, and unknown variance.
Let's suppose we're given $s^2$ (sample variance) and $n$ (sample size).
Is there a way to approximate $P(\bar X>c)$?
I thought of using $\frac{\bar X-\mu}{S/\sqrt{n}}\rightarrow_d N(0,1)$.
However, when I do $P(\bar X>c)=P(\frac{\bar X-\mu}{S/\sqrt{n}}>\frac{c-\mu}{S/\sqrt{n}})$ , I have on the RHS of the inequality a random variable, and not a constant... 
I've seen some writing $P(\bar X>c)=P(\frac{\bar X-\mu}{s/\sqrt{n}}>\frac{c-\mu}{s/\sqrt{n}})$, but then the LHS is not a the r.v. with the t-distribution...
Any help would be appreciated.
 A: If the sample size is moderately large (say $n \ge 100$) and you're 
reasonably sure that the population is normal, then you can use
the sample mean $\bar X$ as an estimate of the population mean $\mu$ and 
the sample standard deviation $S_X$ as an estimate of the population
SD $\sigma,$ then you can just find $P(X > c)$ for $X \sim \mathsf{Norm}(\mu, \sigma).$
However, a population that is sufficiently nearly normal for a valid t test
is not necessarily a good candidate for such a normal-based procedure to
estimate $p = P(X > c),$ especially if $c$ is more than one standard deviation
above the mean. (You're getting into territory where the Central Limit
Theorem converges slowly, or not at all.)

A safer procedure would be to define the event $\{X > c\}$ as a 'Success', and
count the successes $Y$ in $n.$ Then you can use the "Agresti' confidence interval for $p:$
$$ \tilde p \pm 1.96\sqrt{\frac{\tilde p(1 - \tilde p)}{\tilde n}},$$
where $\tilde n = n+4$ and $\tilde p = (Y+2)/\tilde n.$ This procedure
for estimating $p$ is valid even if the population is not normal.
