Rejection sampling from a normal distribution I am running a Monte-Carlo simulation and I sample from various normal distributions. I was just wondering, is there a way by which I can increase the probability of selecting a point from the tails (i.e., $[-5\sigma,-3\sigma]$ and $[3\sigma,5\sigma]$) as compared to the probability associated with the interval $(-3\sigma, 3\sigma)$?
Edit:

 A: Well, after discussion in comments, I think it might be clear enough to hazard an answer to the question.
It seems that a symmetric-about-0 location-mixture of two normals is required, such that the probability of being in (0, 3) is about 0.15 and in (3, 5) is about 0.35 (and the same on the other side of 0).
This can be done; we'll do the positive component and the negative one will simply be the same but with $-\mu$ in place of $\mu$. The positive component should therefore have approximately 0.3 chance of being in (0, 3) about about 0.7 chance of being in (3, 5), since these probabilities will be halved when we select each half with probability 0.5.
Since almost all the probability for this positive part must lie in (0,5), and $\mu$ must exceed 3, $\sigma$ should probably be less than 1 (so that not too much probability is above 5).
Like so:

For a given $\sigma$ we need $P(X<3) = 0.3$, so $P(\frac{(X-\mu)}{\sigma}<\frac{(3-\mu)}{\sigma})=0.3)$, or $\frac{(3-\mu)}{\sigma}=\Phi^{-1}(0.3)$ where $\Phi^{-1}$ is the inverse cdf of the standard normal. Hence $\mu = 3-\sigma\, \Phi^{-1}(0.3)$. 
Calculating in R:
> sig=c(0.6,0.8,1.0); data.frame( sigma = sig, mu = 3-sig*qnorm(0.3) )
  sigma       mu
1   0.6 3.314640
2   0.8 3.419520
3   1.0 3.524401

Check how much probability is above 5 for the $\sigma=1$ case:
> pnorm(5,3.524,1,lower.tail=FALSE)
[1] 0.06997195

That's perhaps a little high, we only get (0.7-0.07)/2 = 0.315 in (3,5). Checking $\sigma=0.8$:
 pnorm(5,3.4195,0.8,lower.tail=FALSE)
[1] 0.02409863

That looks reasonably good, almost 0.34 in (3,5). (The value for $\sigma=0.6$ is 0.349.)
You can use $\mu = 3-\sigma\, \Phi^{-1}(0.3)$ with whatever value of $\sigma$ you prefer, or you can manipulate the equation so that $\mu$ is given and $\sigma$ is calculated.
