0
$\begingroup$

If $X \sim U(2, 8)$

Would it be $$P(X > 1.5 + 4) + P(X <-1.5 +4)$$

$\endgroup$
  • $\begingroup$ Yes, you are correct! $\endgroup$ – SWilliams Jun 14 '15 at 23:13
2
$\begingroup$

You got it. Just follow through with that thought.

$$\begin{split} P(\left| X-4 \right|) > 1.5) &= P(X > 5.5) + P(X < 2.5) \\ &= \frac{2.5}{6} + \frac{0.5}{6} \\ &= \frac12 \end{split}$$

$\endgroup$
  • $\begingroup$ Thanks. Got the same answer as well. much appreciated $\endgroup$ – user2077348 Jun 14 '15 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.