A normal distribution can be completely defined by two parameters - the mean and the standard deviation. Given a normal distribution however, say X, how can I use just the mean and the standard deviation to give me conditional expected values for X<=0 and for X>0? Im guessing the distribution can be standardised to obtain a z-statistic.
statdad
Aug27-08, 04:32 PM
"Given a normal distribution however, say X" - I assume you mean that the variable X has a normal distribution. Are both \mu and \sigma known?
"how can I use just the mean and the standard deviation to give me conditional expected values for X \le 0 and for X>0 ?"
This doesn't make sense to me as it stands. In statistics we take expected values of some function of a random variable - can you elaborate on what it is you seek?
HallsofIvy
Aug27-08, 04:56 PM
If X has normal distribution with mean \mu and standard deviation \sigma then z= (x- \mu)/\sigma has the standard normal distribution. As statdad said, "conditional expected values for X< 0 and X> 0" makes no sense." I might interpret as "suppose X a standard normal distribution, restricted to be larger than 0. What is the the expected value of X?"
That would be
\frac{1}{\sqrt{\pi}}\int_0^\infty x e^{-x^2}dx= \frac{1}{2\sqrt{\pi}}
The general problem, with non-zero mean or standard deviation not 1 would be a much harder integral.