The continuous random variable $X$ is such that \[P(X x) = \begin{cases} 1 & , & x \le 0 \ k

Question

The continuous random variable $X$ is such that $$P(X > x) = \begin{cases} 1 & , & x \le 0 \ k(3 - x)^3 & , & 0 < x < 3 \ 0 & , & x \ge 3 \end{cases}$$ Show that $\mathrm{P}(X > 1) = \frac{8}{27}$. Find the probability density function of $X$, and hence find $\mathrm{E}(X)$.

Accepted Answer

$\mathrm{P}(X>1)=\frac8{27}$. $\mathrm{f}(x)=\frac{(3-x)^2}{9}$ for $0<x<3$, and $0$ otherwise. $\mathrm{E}(X)=\frac34$.

Calculating pdf from a given cdf

Calculating pdf from a given cdf

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