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$F(x)=\begin{cases} 0 & x < 0 \\ x^2 & 0 \le x < \frac{1}{2} \\ \alpha & x = \frac{1}{2} \\ 1 - 2^{-2x} & x > \frac{1}{2} \end{cases} $

  1. Find $\text{Pr}\left (\frac{1}{4} < X \le \frac{3}{4}\right)$.
  2. Find $\alpha$.

Please help, i am a bit confused because of the jump

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1 Answer 1

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  1. $P\left(\tfrac14<X\leq \tfrac34\right)=P\left(X\leq \tfrac34\right)-P\left(X\leq \tfrac14\right)$.
  2. A cumulative distribution function is right-continuous. That is $F(x)=\lim_{y\downarrow x}F(y)$. This enables you to determine $\alpha=F(\tfrac12)$.
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+1 for teaching how to fish instead of etc. –  Did Aug 12 '13 at 19:30
    
thanks bro got it. :D –  FarhanA Aug 12 '13 at 19:31

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