Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

Given the following distribution, (See Below) What is the probability that x<2 given that x<4?

OpenStudy (anonymous):

\[f(x)= \frac{ 2 }{ x ^{3} }\] for x > 1 f(x) = 0 otherwise

OpenStudy (anonymous):

So what I really need is to check my answer with someone... I got an answer but not sure if I did it correctly or not.

OpenStudy (anonymous):

\[P(X<2|X<4)=\frac{P((X<2)\wedge(X<4))}{P(X<4)}=\frac{P(X<2)}{P(X<4)}\] where \[P(X<k)=\int_{-\infty}^kf(x)\,dx=\int_1^k\frac{2}{x^3}\,dx=\frac{1}{x^2}\bigg|_{x=k}^1=1-\frac{1}{k^2}\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!