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Question Number 192377 by Spillover last updated on 16/May/23

$$Theprobabilitydensityfunctionf\left(x\right)$$$$ofavariablexisgivenby$$$$f\left(x\right)=\{\begin{array}{l}kx\sin \pi x0⩽x⩽1\\ 0forallvalueofx\end{array}$$$$Showthatk=\pi anddeducethatmean$$$$andthevarianceofthedistributionare$$$$(1-\frac{4}{{\pi }^2})and\frac{2}{{\pi }^2}(1-\frac{8}{{\pi }^2})$$

Answered by mehdee42 last updated on 16/May/23

$${{\int }_0^{1}kxsin\pi xdx=1\Rightarrow -\frac{k}{\pi }xcos\pi x+\frac{k}{{\pi }^2}sin\pi x]}_0^1=1$$$$\Rightarrow k=\pi ✓$$$$E\left(x\right)={\int }_0^1\pi x^{2}sin\pi xdx=\pi {(-\frac{x^2}{\pi }cos\pi x+\frac{2x}{{\pi }^2}sin\pi x+\frac{2}{{\pi }^2}cos\pi x]}_0^1$$$$\pi (\frac{1}{\pi }-\frac{4}{{\pi }^3})=1-\frac{4}{{\pi }^2}✓$$$$E\left(x^2\right)={\int }_0^1{x}^{3}sin\pi xdx$$$$=\pi {[-\frac{x^3}{\pi }cos\pi x+\frac{3x^2}{\pi }sin\pi x-\frac{6x}{{\pi }^3}cos\pi x+\frac{6}{{\pi }^4}sin\pi x]}_0^1=1-\frac{6}{{\pi }^2}$$$$Var\left(x\right)=E\left(x^2\right)-E^2\left(x\right)=1-\frac{6}{{\pi }^2}-{(1-\frac{4}{{\pi }^2})}^2$$$$=\frac{2}{{\pi }^2}(1-\frac{8}{{\pi }^2})✓$$$$$$

Commented by Spillover last updated on 16/May/23

$$yourright.$$

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