How to find the mean of all z scores above a cutoff I would like to find the mean of all z-scores above a z-score cutoff.  For example, what is the mean of all z scores above a z of 1.96 in a distribution N(0,1).
(My calculus skills are modest.) 
 A: By way of a supplement to Stephan's answer, note that the integration is quite straightforward. Let $\phi$ and $\Phi$ be the density and cdf of the standard normal. 
We need simply notice that the exponent in a standard normal density is $-z^2/2$, which has a derivative of $-z$, and so by the chain rule for differentiation, $\phi'(z) = -z\,\phi(z)$ (with a little experience with integration this stuff becomes fairly easy to spot). Then:
$E(Z|Z>z)=1/(1-\Phi(z))\int_z^\infty u\,\phi(u)\, du =  (1-\Phi(z))\, [-\phi(u)]_z^\infty=\frac{\phi(z)}{1-\Phi(z)}$
A: You are looking for the truncated normal distribution. The Wikipedia page gives explicit formulas for the mean.
In your example,
$$ \alpha=1.96,\quad \beta =\infty.$$
So, as per the Wikipedia page,
$$ \phi(\beta)=0, \quad \Phi(\beta)=1,\quad Z=\Phi(\beta)-\Phi(\alpha)=1-\Phi(\alpha) $$
and since $\mu=0$ and $\sigma=1$, the mean you are looking for is
$$ \mu + \frac{\phi(\alpha)-\phi(\beta)}{Z}\sigma
= \frac{\phi(\alpha)}{1-\Phi(\alpha)}.$$
A quick rejection sample seems to confirm this:
alpha <- 1.96
nn <- 1e6
set.seed(1)
foo <- rnorm(nn)
mean(foo[foo>alpha])
# [1] 2.333103
dnorm(alpha)/(1-pnorm(alpha))
# [1] 2.337835

