Solving Definite Integral Problem with Density Function f(x) = 2e^(-0.25x)

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maxpayne_lhp
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Hello,

Okay so I was given a density function:

[tex]f(x) = 2e^(-0.25x)[/tex]

The problem asks for the value of Pr(X < or = 3)

I first figured out the probability density function first by let

[tex]\int (3,0) k.25e^(-.25x) = 1[/tex]

And figured out that k = .45

and continue solving my problem until i get the final answer is -.212

So, is my solution correct? Did I misunderstand something about k?

Thanks much!

NN
 
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Sorry, my LaTeX code looks bad.
 
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[tex]k\int_{0}^{\infty} e^{-0.25 x}{}dx =1[/tex]

I don't know about the value of k, whether it's correct or not, guess not...

Daniel.
 
What do you mean, you were "given a density function:
[itex]f(x)= 2e^{-.25x}[/itex]"? You seem to think that you need to find a multiplier k so that the "total integral" (is that supposed to be an integral from 0 to 3?) is 1. If that were the case then you wouldn't be asked for the probability that x is between 0 and 3: it would be 1!

Dextercioby, on the other hand, seems to think that you need to find k such that the integral from 0 to infinity is 1.

I see that we are already told that the function is [itex]2e^{-.25x}[/itex]. That is, that the "k" is 2, but that we are not given an interval over which this is to be the probability density.
 
Daniel, yeah thanks for the code... i meant

[tex]k\int_{0}^{3} e^{-0.25 x}{}dx =1[/tex]

HallsofIvy, Uh... Thats the same word from the problem in the book... and yes, thats the integral from 0 to 3... I am not sure I know about the total integral...

Sorry about the confusion.
 
Then please state the problem exactly, word for word. So far you have told us that the density function is [tex]f(x) = 2e^{-0.25x}[/tex]
that [tex]k\int_{0}^{3} e^{-0.25 x}{}dx =1[/tex]
and that the problem asks for the probability that x is less than or equal to 3. I'm afraid none of that makes much sense to me. What exacly is the probability function and over what range is it defined?