Question attached as an image. It is # 6. Any help would be awesome, thanks

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- Dec 3rd 2009, 10:46 AMbabbaganducumulative distribution
Question attached as an image. It is # 6. Any help would be awesome, thanks

- Dec 3rd 2009, 01:39 PMFocus
Could you please post an attempt? I don't mean to be rude but you are just posting a load of questions, for which it seems you shown no effort to do.

Here if a tip to get you started $\displaystyle \mathbb{P}(f(X)\leq a)=\mathbb{P}(X\leq f^{-1}(a))$. - Dec 5th 2009, 09:51 AMbabbagandu
Thank you for the tip, but I am still a little confused. Could I have another hint maybe?

- Dec 6th 2009, 02:46 PMFocus
Use this;

in conjunction with the fact that $\displaystyle e^{\log X}=X$. When you have, you know what distribution X has, so you can explicitly write out the value.

If you want the hint more explicitly: $\displaystyle \log X \leq a$ take exponential of both sides.