anonymous
  • anonymous
The probability density function\[f(x)=\left(\begin{matrix}c(4-x^2) \\ 0\end{matrix}\right)\] C = constant = 3/32\[-2 \le x \le 2\]otherwise. Find \[P(|x| \ge 1).\]
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

ash2326
  • ash2326
\[P(|x|\ge1) = P(-\infty\le x\le -1)+P(1\le x \le\infty)\] \[P(|x|\ge1) = \int_{\infty}^{-1}f(x)dx+\int_{1}^{\infty}f(x) dx\] since f(x) in non zero for \(-2\le x \le 2\) we'll have \[P(|x|\ge1) = \int_{-2}^{-1}f(x)dx+\int_{1}^{2}f(x) dx\] here \[f(x)= \frac{3}{32} (4-x^2)\] can you do it now?
anonymous
  • anonymous
I understand it, but I don't know how to remove |x| to make it x. Is it just adding the - integrals to the positive integrals? i.e. \[\int\limits_{-1}^{2} and \int\limits_{2}^{1}?\]
anonymous
  • anonymous
*-2

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ash2326
  • ash2326
You need not remove |x|, that we have already included. Substitute the value of f(x) and evaluate the two integrals
anonymous
  • anonymous
Ok, yes, I understand how to do the integration from then on. But I'm not sure how you got the -2 < x < -1 and 1 < x < 2 in the first place?
ash2326
  • ash2326
see we want probability in the range \(|x|\ge 1\) this is equivalent to \(-\infty \le x\le 1\) and \( 1 \le x\le \infty\) therefore I split the integral in two ranges. f(x) is non zero in the range \(-2 \le x\le 2\), therefore it boils down to two integrals from -2 to -1 and from 1 to 2
anonymous
  • anonymous
Ah Ok. Thank you!
anonymous
  • anonymous
Do you think you could help me with one more example?
anonymous
  • anonymous
I mean, I have another question
ash2326
  • ash2326
I'll try, post it as a new question

Looking for something else?

Not the answer you are looking for? Search for more explanations.