Probability that half of a sample from a normal distribution will be above a threshold Suppose we have $n$ samples drawn from a standard normal distribution.
What is the probability that at least half $(\frac{n}{2})$ of these samples are above some threshold above that mean, say, above $1.5\sigma$?
I understand that the probability of a single draw from a normal distribution is $1 - \Phi(1.5)$ (where $\Phi$ is the cumulative distribution function).
My immediate thought is to suggest that the probability of the first $(n/2)$ draws being above the threshold is the product of this, i.e.:$\Pi_0^{n/2} 1 - \Phi(1.5)$. And then I need to take into account all the permutations in which this can occur, ie. every odd draw might be above $1.5\sigma$, or alternatively, all $n$ draws might be above $1\sigma$, which would involve a summation over the permutations.
Is this correct?
 A: Your general idea is sort of along the right lines but it looks like you didn't quite get there. 
Consider the case where $n$ is even and let's define $m=n/2$. Take that the probability of exceeding the threshold $t$ is $p=1-\Phi(t)$. Let $X$ be the number of values that exceed the threshold; and assume that the events are independent. Then the distribution of $X$ will be binomial$(n,p)$ (as suggested by Hugh in comments).
So the probability of observing exactly $x$ observations over the threshold will be given by the binomial pmf $P(X=x)={n\choose x} p^x(1-p)^{n-x}$ (there's that number of combinations you suggested). 
Consequently the probability of at least $m$ observations above the threshold is the same as $m$ or fewer observations below it, so the answer can be obtained from the cdf for a $\text{binomial}(n,p)$. Similar calculations occur for the case where $n$ is odd.
Many packages will evaluate the binomial cdf for you, though for small $n$ it's feasible to do it by hand and for large $n$ and $p$ not close to $0$ or $1$ you can use a normal approximation.
