If the probability density of a random vribale is given by :

f(x)= x for 0

Question

If the probability density of a random vribale is given by :

f(x)= x               for 0<x<1
       2-x            for 1<=x<2
        0              elsewhere

find the probabilities that a random variable having this probability density will take on a value

(a)between .2 and .8?

(b) between .6 and 1.2?

anonymous · Answer

I already did the first one here: http://openstudy.com/users/jerjason#/updates/509d2b9fe4b0ac7e51942b75 I did this example in class and the answers my professor gave me for (a) was .3 for (a) and .5 for (b), my question is why am I only evaluating the integrals for f(x) = x and not f(x)=2-x as well?

anonymous · Answer

$$\int\limits_{.6}^{1.2}x dx = 1/2(x^2)]_{.6}^{1.2}=.54$$
is what I got for (b) and rounded it is .5.

sirm3d · Answer

$$\large \int\limits_{0.6}^{1.2}f(x) dx=\int\limits_{0.6}^{1}f(x) dx+\int\limits_{1}^{1.2}f(x) dx$$

sirm3d · Answer

use $\large f(x)=x$ for the first integral and $\large f(x)=2-x$ for the second integral

sirm3d · Answer

as to your question of why only $f(x)=x$ in (a), the function clearly states that $$\large f(x)=\begin{cases} x & 0<x<1,\2-x & 1 \leq x< 2,\0 & \text{elsewhere} \end{cases}$$

sirm3d · Answer

you must use $f(x)=x$ when $0 < x < 1$ since the interval given in (a) is a subset of this interval.

anonymous · Answer

Alright, I understand now. Thanks.