Let X be a uniform variate defined on (-k,k). Determine k so that
P(|X|<1) = P(|X|>2)
Hello,
Draw a line where you represent -2,-1,0,1,2 and represent the areas where |X|<1 and |X|>2 and you'll see that for the areas to be equal, the domain has to stop to -3 and 3.
More formally... :
$\displaystyle P(|X|<1)=\frac{1}{2k} \int_{-1}^1 ~dt=\frac 1k$
$\displaystyle P(|X|>2)=\frac{1}{2k}\left(\int_{-k}^{-2} ~dt+\int_2^k ~dt\right)=\frac{1}{2k}\cdot (-2+k+k-2)=\frac{k-2}{k}$
Thus we must have $\displaystyle \frac 1k=\frac{k-2}{k} \Rightarrow 1=k-2\Rightarrow k=3$
boldeagle : what's the point giving the solution without further information ? It doesn't help