Is the quotients of a group of triangular distributed numbers still following a triangular distribution? I have a group of numbers (about 10000 numbers) between 0.8 and 1.0 which follows simple triangular distribution (for example, lower limit: 0.8, upper limit: 1.0, mode: 0.9).
If I divide 2 by each number from this group, I got the quotients forming the second group of numbers which follows another distribution (i.e. lower limit: 2.0, upper limit: 2.5, mode: 20/9).
Here is my question: is the second distribution still a triangular distribution? If not, then how can I know quantitatively what kind of distribution it is?
I will appreciate very much for your kind answers!
 A: The mgf of the triangular distribution is $$M_{X}(t) = \frac{2(b-c)e^{at}-(b-a)e^{ct}+(c-a)e^{bt}}{(b-a)(c-a)(b-c)t^{2}}$$
where $a,b,c$ are the lower limit, upper limit, and mode respectively. Note that this is an alternate way of characterizing the probability distribution function. Replacing $a := 2/a$, $b:= 2/b$ and $c:= 2/c$ still produces a triangular distribution.
A: We work with your numbers $0.8$ and $1$. So the base of the triangle has length $0.2$. To make the area $1$, the height of the triangle needs to be $10$. 
We are dealing with a random variable $X$ whose density function we could readily find. In order to follow what is coming, you need to draw a picture of that triangle. 
Now we look at the random variable $Y=\dfrac{2}{X}$, and we want a description of the cumulative distribution function of $Y$, or maybe its density function, you did not specify. We go after the cumulative distribution function $F_Y(y)$. You can then differentiate if you need the density function. The distribution will not be triangular, though it it looks roughly triangular.
The effective range of $Y$ is, as you pointed out, $2$ to $2.5$. Let $F_Y(y)$ be the cumulative distribution function of $Y$, that is, $\Pr(Y\le y)$. It is clear that $F_Y(y)=1$ if $y\ge 2.5$, and $F_Y(y)=0$ for $y\le 2$. We now need to do some work to find $F_Y(y)$ for $y$ in the interesting interval, $2$ to $2.5$. We have
$$F_Y(y)=\Pr(Y\le y)=\Pr\left(
\frac{2}{X} \le y\right))=\Pr\left(X\ge \frac{2}{y}\right).$$
First deal with the case $y \ge \frac{2}{0.9}$. Then $\frac{2}{y}\le 0.9$. We want the probability that $X\ge \frac{2}{y}$. This is the area of the part  of the triangle that is to the right of $\frac{2}{y}$.  
The area of the part of the triangle that is to the left of $\frac{2}{y}$ is easier to find directly. This is a triangle with base $\frac{2}{y}-0.8$. The height of the triangle is $100(\frac{2}{y}-0.8)$. So the area of the triangle is $50(\frac{2}{y}-0.8)^2$, and therefore for $y \ge \frac{2}{0.9}$ (but $\le 2.5$) we have
$$F_y(y)= 1-50\left(\frac{2}{y}-0.8\right)^2.$$
Of course I don't trust my calculation. So let's check what happens at $y=\frac{2}{0.9}$, where the answer should be $0.5$. It is! The answer should be $1$ at $y=2.5$. Yup.
Now we need to deal with $y$ between $2$ and $\frac{2}{0.9}$. The probability that $Y\le y$ is the area of the part of our original triangle which is to the right of $\frac{2}{y}$. The base of this triangle is $1-\frac{2}{y}$. The height turns out to be $100$ times that, so for $2\lt y\le \frac{2}{0.9}$ we have
$$F_Y(y)=50\left(1-\frac{2}{y}\right)^2.$$
We can differentiate to find the density function $f_Y(y)$.  
If you graph the resulting density, the distribution of $Y$ will probably look triangular to the casual eye. It isn't, we have the exact distribution above. But triangular is likely a good enough approximation  for your purposes. If you went say $0.15$ to $1$ instead of $0.8$ to $1$, the departure from triangularity would be significant. Here it may not be.
Added: Explicitly, by differentiating the expressions for the cdf above, we find that $f_Y(y)$, the density function of $Y$,  is in the interval $(2,2.5)$  given by:
$f_Y(y)=\dfrac{200}{y^3}(y-2)$ for $2\le y \le \frac{2}{0.9}$, and
$f_Y(y)=\dfrac{200}{y^3}(2-0.8y)$ for $\frac{2}{0.9} \lt y\lt 2.5$.
