In working out a recent blog post, I had cause to find the probability, in a standard normal distribution, of $z 1, I was stumped. Could I get anywhere close mentally? A good question.

Starting from the definition of the normal distribution, $p(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2}$, I figured that modelling the remainder of the graph as a triangle might be fruitful.

For $x = -23$, we’re looking at the height of the graph, $p(-23) = \frac{1}{\sqrt{2\pi}} e^{-\frac{-529}{2}}$. I’m not going to work that out yet.

What’s the gradient of the curve there? $\diff{p}{z} =…

Continue reading at:

http://ift.tt/2fvntXu

### Like this:

Like Loading...

*Related*