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Let $$F(x) = 1 − \Bbb e ^{-x^3}; x > 0$$ be the cumulative distribution function of a continuous random variable $X$.

(a) Find the probability density function of $X$.

(b) Find the value of $c$ such that $P(X < c) = 0.5$.

For (a), do I just take the derivative of $F(x)$ to find $f(x)$? (in which case the pdf is $3x^2e^{-x^3}$ )

I haven't tried for (b) since I'm not sure on (a).

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    $\begingroup$ a) is correct! Any ideas on b)? $\endgroup$ – user190080 Jun 28 '15 at 14:33
  • $\begingroup$ as I understand the problem, c can take on any value greater then $1/2$ since $P(X<c) = 1/2$ $\endgroup$ – Joz Jun 28 '15 at 14:48
  • $\begingroup$ $\int_0^c 3x^2e^{-x^3} dx=\frac{1}{2}$? $\endgroup$ – Roland Jun 28 '15 at 14:50
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    $\begingroup$ not quite sure what you mean, but you better check the definition of a cdf, this is $F(c)=P(X\le c)$ $\endgroup$ – user190080 Jun 28 '15 at 14:50
  • $\begingroup$ this is what I show in my notes: $$ P(X <c) = F(c) − lim F(x).$$ $\endgroup$ – Joz Jun 28 '15 at 15:32
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(a) Indeed, the probability density function of $X$ can be obtained by differentiating the cumulative distribution of $X$. So your answer $f(x)=F'(x)=3x^2e^{-x^3}$ is correct.

(b) Here you simply use the definition of cumulative distribution function, $P(X<c)=F(c)$, so \begin{eqnarray} 0.5 &=& F(c) = 1-e^{-c^3}\\ e^{-c^3}&=&1/2 \\ c &=& (\ln2)^{1/3} \end{eqnarray}

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