p.d.f.

**antman** · April 17th 2009, 11:30 AM

The p.d.f. of time X to failure of an electronic component is $f(x)=2x/1000^{2}*e^{-(x/7)^{2}},$ $0<x<\infty$

a) Compute P(X>2000)
b) Determine the 75th percentile, $\pi_{0.75}$ , of the distribution
c) Find the 10th and 60th percentiles, $\pi_{0.10}$ and $\pi_{0.60}$

**Moo** · April 17th 2009, 12:58 PM

Hello,

Originally Posted by antman

The p.d.f. of time X to failure of an electronic component is $f(x)=2x/1000^{2}*e^{-(x/7)^{2}},$ $0<x<\infty$

a) Compute P(X>2000)

By definition of the cumulative density function, this is $\int_{2000}^\infty f(x) ~dx$

b) Determine the 75th percentile, $\pi_{0.75}$ , of the distribution

Find t such that $\int_0^t f(x) ~dx=0.75$

c) Find the 10th and 60th percentiles, $\pi_{0.10}$ and $\pi_{0.60}$

Same as b)

**antman** · April 23rd 2009, 07:22 AM

For part a, I got an answer of $e^{-4}$ or 0.0183. I'm not sure how to solve for t in parts b and c. My book doesn't show this way of doing the problem or any other way for that matter. Thank you.

**antman** · April 24th 2009, 11:04 AM

For part b, I set the equation up as $\int$ with limits $0\le t \le .75$ with $f(t)=\frac{2t}{1000^{2}}e^{-(\frac{t}{1000})^{2}} dt$ but got an answer of 0.000000564, which seems too small to be correct. I tried it also with 75 as my upper limits and got 0.00561, which doesn't seem right either. Can anyone help me figure out what I am doing wrong? Thank you.

**mr fantastic** · April 24th 2009, 03:00 PM

Originally Posted by antman

For part b, I set the equation up as $\int$ with limits $0\le t \le .75$ with $f(t)=\frac{2t}{1000^{2}}e^{-(\frac{t}{1000})^{2}} dt$ but got an answer of 0.000000564, which seems too small to be correct. I tried it also with 75 as my upper limits and got 0.00561, which doesn't seem right either. Can anyone help me figure out what I am doing wrong? Thank you.

75th percentile means finding the value of x such that Pr(X < x) < 0.75. So the unknown x is an integral terminal and 0.75 is the value of the integral.

**antman** · April 25th 2009, 08:48 AM

When I take the integral, I get the equation $-e^{-\frac{1}{1000000}x^{2}}=0.75$ . Then taking the ln of both sides leaves $-\frac{1}{1000000}x^{2}=-.287682$ . Multiplying both sides by -1000000 gives $x^{2}=287682.0725$ , so x=536.36? Is this how you would solve for x? I don't know what to do with the -e. That's where I got stuck and was looking for an alternative way to solve the problem. Hopefully I have the right idea. Thank you.

**mr fantastic** · April 25th 2009, 04:19 PM

Originally Posted by antman

When I take the integral, I get the equation $-e^{-\frac{1}{1000000}x^{2}}=0.75$ . Then taking the ln of both sides leaves $-\frac{1}{1000000}x^{2}=-.287682$ . Multiplying both sides by -1000000 gives $x^{2}=287682.0725$ , so x=536.36? Is this how you would solve for x? I don't know what to do with the -e. That's where I got stuck and was looking for an alternative way to solve the problem. Hopefully I have the right idea. Thank you.

Check: If you try integrating between 0 and your answer, you should get 0.75 if it's correct. I suggest using technology to make the job easier.

**antman** · April 26th 2009, 08:48 AM

By the process of elimination, I got approximately 1177.41 but I don't know how to get there the right way. Does anyone have any ideas? Thank you.

**antman** · April 26th 2009, 10:23 AM

I got it! Thank you for your help. I integrated incorrectly but finally am able to get the answer through integration by hand. Thanks.

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