3
$\begingroup$

Suppose I have the following probability density function.

$$f(x) = \begin{cases}ce^{-\frac x{200}}&\mathrm{\ if\ } 0

How do I find $c$? Currently, I have the following.

$$\int_0^\infty ce^{-\frac x{200}}dx = 1$$

$$\lim_{x\to\infty} -200ce^{-\frac x{200}} + 200c = 1$$

$$c = \frac1{200}$$

Is this right?

2 Answers 2

3

Yes, that is completely correct

2

So we want to choose $c$ so that $$\int_0^\infty ce^{-x/200}\, dx=1.$$

An antiderivative of $ce^{-x/200}$ is $-200ce^{-x/200}$. So our definite integral is $200c$. It follows that $c=1/200$.

  • 0
    Why would you assume when the conditions are listed?2012-09-12
  • 0
    No, the conditions were listed.2012-09-12
  • 0
    @idealistikz: Yes, you are right, it wasn't in TeX so I missed it.2012-09-12