# Probability pdf

• Jul 3rd 2008, 10:20 PM
amor_vincit_omnia
Probability pdf
Hey guys, here's a problem I'm stuck on:

The lifetime of X (years) has pdf:

f(x) = 2x/(theta²) *e^ -(x/theta)² , when... 0<x<infinity

If P(X>5) = 0.01, determine theta.

--------------

So I thought, P(x>5) = 1-P(x<5)

so, 1- integral (from 0-->5) f(x) = 0.01

Is this the right logic? If so, the problem becomes evaluating that integral.
Can someone help me? It's been two years since Calculus, and I'd appreciate it.
• Jul 3rd 2008, 10:38 PM
flyingsquirrel
Hi
Quote:

Originally Posted by amor_vincit_omnia
The lifetime of X (years) has pdf:

f(x) = 2x/(theta²) *e^ -(x/theta)² , when... 0<x<infinity

If P(X>5) = 0.01, determine theta.

--------------

So I thought, P(x>5) = 1-P(x<5)

so, 1- integral (from 0-->5) f(x) = 0.01

Is this the right logic? If so, the problem becomes evaluating that integral.

$\displaystyle 1-\int_0^5 \frac{2x}{\theta^2}\mathrm{e}^{-\frac{x^2}{\theta^2}}\,\mathrm{d}x=0.01$

What you've done is correct. To evaluate the integral, notice that $\displaystyle \left(-\frac{x^2}{\theta^2}\right)'=-\frac{2x}{\theta^2}$ : you may try substituting $\displaystyle u=\frac{x^2}{\theta^2}$.
• Jul 3rd 2008, 11:13 PM
amor_vincit_omnia
okay, for the final answer i'm getting

1 - {1/ [e^(25/θ²)]}

anyone else? Thanks for the help, wish I could find all of my old Cal notes!
• Jul 3rd 2008, 11:19 PM
flyingsquirrel
Quote:

Originally Posted by amor_vincit_omnia
okay, for the final answer i'm getting

1 - {1/ [e^(25/θ²)]}

That's it :)