Relationship between Bernoulli and Normal CDF Is there any relationship between the draw from a Bernoulli with parameter $p$ and the Normal CDF. Specifically is the condition $p>\Phi(x)$, where $x$ is drawn from the standard normal distribution, equivalent to a draw from Bernoulli$(p)$ ?
 A: 
Is there any relationship between the draw from a Bernoulli with parameter p and the Normal CDF. 

Yes, but usually it's not terribly interesting on its own. 
There are some contexts where relationships between variables might make use of such a relationship (such as a probit regression for binomial data, or in some correlation measures involving one or both variables being binary).

Specifically is the condition p > cdf(x) where x is drawn from the standard normal distribution equivalent to a draw from Bernoulli(p) ?

If the cdf you mean is the standard normal cdf, $\Phi$, then $\Phi(X)$ is a (continuous) uniform on (0,1). See Probability Integral Transform.
You can generate Bernoulli random variables with parameter $p$ by generating uniform and comparing the uniform against $p$... however, the way it works is if $U>p$ you generate a 0; if $U\leq p$ you generate a 1. This corresponds to the inverse-cdf method of random number generation (at least, it is with a minor algebraic fiddle).
So if you're using this as a source of Bernoulli random numbers, it's a very inefficient way of proceeding: generating normal random numbers, turning your normal values into uniform values and using uniforms to get the Bernoulli random numbers. You might as well just generate uniforms directly and avoid a lot of mucking about.
