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

maxpayne_lhp
Messages
35
Reaction score
0
Hello,

Okay so I was given a density function:

f(x) = 2e^(-0.25x)

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

I first figured out the probability density function first by let

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

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
 
Physics news on Phys.org
Sorry, my LaTeX code looks bad.
 
Click this picture to see where your typing went wrong

k\int_{0}^{\infty} e^{-0.25 x}{}dx =1

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:
f(x)= 2e^{-.25x}"? 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 2e^{-.25x}. 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

k\int_{0}^{3} e^{-0.25 x}{}dx =1

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 f(x) = 2e^{-0.25x}
that k\int_{0}^{3} e^{-0.25 x}{}dx =1
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top