# Easy way to solve a symmetric triangular distribution, without using integrals

Suppose $P_\text{tax}$ has a symmetric triangular distribution on $[1.2,1.9]$. Please answer:

a) The probability that $P_\text{tax}< 1.6$.

I have the following answer without the use of integrals:

triangle base $= 1.55-1.2$

triangle height $= 2/(1.9 - 1.2)$.

trapezoid height $= 1.6-1.55$

trapezoid 'bigger' base $= 2/(1.9-1.2)$

trapezoid 'smaller' base $= -4(1.6-1.9)/(1.2-1.9)^2$

Hence, the probability that $P_\text{tax}< 1.6$ is $0.5+0.2653 = 0.7653$.

-
What is a Ptax? It sounds like some kind of fossilized mosquito... –  Mariano Suárez-Alvarez Nov 9 '11 at 15:28
Ptax is an exchange rate –  jvb Nov 9 '11 at 15:31
I assume that by symmetric triangular distribution we mean that the density function has shape an isosceles triangle base the interval from $1.2$ to $1.9$. Then the given answer is wrong. –  André Nicolas Nov 9 '11 at 15:45

It looks as if the intended geometry is the following. We have an isosceles triangle with base the interval $[1.2, 1.9]$ and area $1$. We want to find the area of the part of the triangle to the left of the line $x=1.6$.

There are various ways of computing that area. The first step, as always, is to draw a good diagram. To me, the region to the left of $x=1.6$ looks kind of ugly, whil the region to the right of $x=1.6$ looks nice. So we will compute the area to the right of $x=1.6$, and subtract this area from $1$.

We want to find the area of the triangle to the right of $x=1.6$. This triangle has base $0.3$. The triangle from $x=1.55$ on has base $0.35$, area $1/2$, and is similar to our target triangle. Recall that scaling the linear dimensions of a figure by the scale factor $t$ multiplies area by $t^2$.

Thus the area of our target triangle is $$\frac{1}{2}\left(\frac{0.3}{0.35}\right)^2. \qquad(\ast)$$

We can now do the arithmetic. Subtract the area $(\ast)$ from $1$. We get about $0.632653$.

Alternately, and more painfully, we can find the area of the trapezoid from $x=1.55$ to $x=1.6$, and add $1/2$ to the area of this trapezoid. That is the idea that was used in the solution mentioned in the post.

The full triangle has base $0.7$ and area $1$, so it has height $2/(0.7)$. That gives one "base" of the trapezoid. We find the other base of the trapezoid. By similar triangles, this other base is $(2/(0.7))(0.3/0.35)$. So the average of the two bases of the trapezoid is $$\frac{1}{0.7}\left(1+\frac{0.3}{0.35}\right).$$ Multiply this average by the "height" $0.05$ of the trapezoid. We get about $0.132653$. Finally, add $0.5$.

Comment: You computed the "other base" correctly. So what went wrong is in the part you did not give detail about, finding the area of the trapezoid given the two bases and the height.

-