## anonymous 4 years ago 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 $P(|x| \ge 1).$

1. 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?

2. 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. $\int\limits_{-1}^{2} and \int\limits_{2}^{1}?$

3. anonymous

*-2

4. ash2326

You need not remove |x|, that we have already included. Substitute the value of f(x) and evaluate the two integrals

5. 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?

6. 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

7. anonymous

Ah Ok. Thank you!

8. anonymous

Do you think you could help me with one more example?

9. anonymous

I mean, I have another question

10. ash2326

I'll try, post it as a new question