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 $$P(|x| \ge 1).$$

Question

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 $$P(|x| \ge 1).$$

ash2326 · Answer

$$P(|x|\ge1) = P(-\infty\le x\le -1)+P(1\le x \le\infty)$$
$$P(|x|\ge1) = \int_{\infty}^{-1}f(x)dx+\int_{1}^{\infty}f(x) dx$$
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 · Answer

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 · Answer

*-2

ash2326 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

Ah Ok. Thank you!

anonymous · Answer

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

anonymous · Answer

I mean, I have another question

ash2326 · Answer

I'll try, post it as a new question