MAD & Standard Deviation Am I correct in saying that you can only use the formula for approximating the standard deviation from MAD, i.e 
$$\text{SD } = K \times \text{ MAD }$$
if you know the actual probability distribution?
I have seen in a previous post that $K \approx 1.48$ for the normal distribution.
What is it for the Poisson distribution? 
I have also seen an expression for $K = 1.25$.
 A: I'll assume that you're interested in the median absolute deviation from the median.
The ratio of SD/MAD is different for each distribution.
All normal distributions have the same ratio, but for Poisson distributions the ratio depends on the Poisson parameter. 
Note that if your data are really Poisson, an actual confidence interval for the population sd ($\sqrt{\lambda}$) would be best obtained by generating a corresponding interval for $\lambda$ (i.e. based on the sample mean) and then taking the square roots of the limits. As such I'll focus on using $k\times$MAD as a quick approximation for the sample SD though many of the comments would carry over to using it to estimate $\sqrt{\lambda}$.
It turns out that for large samples the asymptotic value for $k$ (that for the normal) works pretty well for a wide range of cases for $\lambda$, as long as the sample MAD is not too small.
I undertook a small simulation study.
For example, when I simulated samples from a very wide range of $\lambda$ values for n=100, I found that $k=1.48$ was pretty good, and that - as long as the MAD was at least 2.5 - the sample standard deviation tended to be within (0.8,1.3)$\times$1.48$\times$MAD more than 90% of the time (the exact performance depends on how you distribute your $\lambda$ values but didn't seem highly sensitive to it). I'd probably lean toward making $k$ just a little smaller but the variation about it is pretty large so it really doesn't matter all that much.
[$k=1.48$ was also reasonable for estimating the population SD, but it looks like the suggested bounds would need to be somewhat wider.]
However for small sample sizes (e.g. at n=10) the MAD was very variable and didn't perform well; the standard deviation will almost always exceed the MAD, so the MAD itself provides a good lower bound, but unless the MAD was quite large the sample SD might reasonably easily be 5+ times as large as the MAD, and while there isn't really any good choice of $k$, you would probably want a smaller value than 1.48 when the MAD is large enough to try it. 
In short: for large samples and MAD at least 2.5, the normal value ($k=1.48$) seems to be adequate. For small samples or MAD below 2.5, I wouldn't be keen to use MAD this way, but you can reasonably assume that the sample SD is at least $1\times$MAD
