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Question Number 145675 by syamilkamil1 last updated on 07/Jul/21

$$theprobabiltydensityfunction$$ $$isknownasfollows:$$ $$f\left(x\right)=\{{}_{0,xother}^{{cx}^3,0<x<4}$$ $$defineP(1<x<2)!$$ $$$$ $$$$ $$$$ \n

Answered by imjagoll last updated on 07/Jul/21

$${\int }_0^4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=1$$ $$\Rightarrow {\int }_0^4{\mathrm{cx}}^3\mathrm{dx}=1$$ $$\Rightarrow \frac{1}{4}\mathrm{c}\left(4^4\right)=1\Rightarrow \mathrm{c}=\frac{1}{4^3}$$ $$\mathrm{P}(1<\mathrm{x}<2)={\int }_1^2\frac{1}{64}{\mathrm{x}}^3\mathrm{dx}$$ $$=\frac{1}{256}(16-1)=\frac{15}{256}$$ $$$$

Answered by puissant last updated on 07/Jul/21

$${\int }_{\mathbb{R}}{\mathrm{cx}}^3\mathrm{dx}=1\Rightarrow \mathrm{c}{\int }_0^4{\mathrm{x}}^3\mathrm{dx}=1$$ $$\Rightarrow \mathrm{c}{\left[\frac{{\mathrm{x}}^4}{4}\right]}_0^4=1\Rightarrow 64\mathrm{c}=1$$ $$\mathrm{c}=\frac{1}{64}$$ $$\mathrm{P}(1<\mathrm{x}<2)=\frac{1}{64}{\int }_1^2{\mathrm{x}}^3\mathrm{dx}=\frac{1}{256}(16-1)$$ $$\mathrm{P}(1<\mathrm{x}<2)=\frac{15}{256}=0,059$$ $$\mathrm{P}(1<\mathrm{x}<2)=5,9\mathrm{\%}..$$

Answered by Dwaipayan Shikari last updated on 07/Jul/21

$$f\left(x\right)={cx}^3$$ $${\int }_0^4{cx}^{3}dx=1\Rightarrow c=\frac{1}{64}$$ $$P(1<x<2)=\frac{1}{64}{\int }_1^2{x}^{3}dx=\frac{1}{256}(2^4-1)=\frac{15}{256}$$

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