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If we let $X$ be a random variable with c.d.f. $F(X)=x^3$ for $0 \le x \le 1$ as my cumulative distribution function and I need to find $P(X\ge \frac12)$ is this correct? $$\int_{1/2}^1 x^3dx=\frac{15}{64} $$ or is the correct way to find the probability this way: $$1-\int_0^{\frac12} x^3dx=\frac{64}{65}$$ I think it is the second way because by looking at a graph it would make sense, however, it seems that this probability is too high for it to be correct.

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You have $P(X\ge 1/2) = 1 - P(X\le 1/2) = 1 - 1/8 = 7/8.$ Don't integrate again. The density is $f_X(x) = 3x^2$, $0\le x \le 1$.

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So for a cumulative density function were simply "plugging in" the value we want into $F(X)$? –  TheHopefulActuary Nov 7 '12 at 0:46
    
what makes you think $0<x<1$? –  Alex Nov 7 '12 at 0:49
    
@ncmathsadist i appologize, the problem states $F(x)= x^3$ for $0 \le x \le 1$ –  TheHopefulActuary Nov 7 '12 at 0:55
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Just an addition to ncmathsadist's answer. Nothing in your question says about the support of the distribution, i.e. on which set $X$ is defined. Hence what you are doing, i.e. $\int_{\frac{1}{2}}^{1} F(x)dx$ does not make any sense, but what you know for sure is that $F(\infty)=1$ and $F \big(\frac{1}{2} \big)=\frac{1}{8}$ and this is all you need.

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