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Determine $k$ so that the following is a valid pdf for $X$ :
$f(x)=k / \sqrt[3]{x}, \quad 0<x<4$
Then find $E(X), P(X<2)$, and $\operatorname{Var}(X)$
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We set the definite integral over the support equal to 1 to find the value of $k$ :
$1=\int_{0}^{4} k x^{-1 / 3} d x=\left.k \frac{3}{2} x^{2 / 3}\right|_{0} ^{4}=\frac{k 3 \sqrt[3]{16}}{2}=k 3 \sqrt[3]{2}$
So $k=1 / 3 \sqrt[3]{2}$
$\begin{gathered} E(X)=\int_{0}^{4} \frac{1}{3 \sqrt[3]{2}} x^{2 / 3} d x=\left.\frac{1}{3 \sqrt[3]{2}} \frac{3}{5} x^{5 / 3}\right|_{0} ^{4}=\frac{1}{5 \sqrt[3]{2}} \sqrt[3]{4^{5}}=\frac{\sqrt[3]{512}}{5}=\frac{8}{5}=1.6 \\ P(X<2)=\int_{0}^{2} \frac{1}{3 \sqrt[3]{2}} x^{-1 / 3} d x=\left.\frac{1}{3 \sqrt[3]{2}} \frac{3}{2} x^{2 / 3}\right|_{0} ^{2}=\frac{1}{\sqrt[3]{4}} \\ E\left(X^{2}\right)=\int_{0}^{4} \frac{1}{3 \sqrt[3]{2}} x^{5 / 3} d x=\left.\frac{1}{3 \sqrt[3]{2}} \frac{3}{8} x^{8 / 3}\right|_{0} ^{4}=\frac{1}{8 \sqrt[3]{2}} \sqrt[3]{4^{8}}=\frac{\sqrt[3]{32768}}{8}=\frac{32}{8}=4 \end{gathered}$
So $\operatorname{Var}(X)=E\left(X^{2}\right)-E(X)^{2}=4-1.6^{2}=1.44$
by Diamond (38,951 points)

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