Question Number 145675 by syamilkamil1 last updated on 07/Jul/21
$${the}\:{probabilty}\:{density}\:{function} \\ $$$${is}\:{known}\:{as}\:{follows}\:: \\ $$$${f}\left({x}\right)\:=\:\left\{_{\mathrm{0}\:\:\:,\:{x}\:{other}} ^{{cx}^{\mathrm{3}} \:\:\:,\:\mathrm{0}\:<\:{x}\:<\:\mathrm{4}} \right. \\ $$$${define}\:{P}\left(\mathrm{1}\:<\:{x}\:<\:\mathrm{2}\right)! \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by imjagoll last updated on 07/Jul/21
$$\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \mathrm{cx}^{\mathrm{3}} \:\mathrm{dx}\:=\:\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\mathrm{c}\left(\mathrm{4}^{\mathrm{4}} \right)\:=\:\mathrm{1}\:\Rightarrow\mathrm{c}\:=\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} } \\ $$$$\mathrm{P}\left(\mathrm{1}<\mathrm{x}<\mathrm{2}\right)=\int_{\mathrm{1}} ^{\:\:\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{64}}\mathrm{x}^{\mathrm{3}} \:\mathrm{dx}\: \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{256}}\left(\mathrm{16}−\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{256}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by puissant last updated on 07/Jul/21
![∫_R cx^3 dx=1 ⇒ c∫_0 ^4 x^3 dx=1 ⇒ c[(x^4 /4)]_0 ^4 =1⇒ 64c=1 c=(1/(64)) P(1<x<2)=(1/(64))∫_1 ^2 x^3 dx =(1/(256))(16−1) P(1<x<2)=((15)/(256)) = 0,059 P(1<x<2)= 5,9%..](https://www.tinkutara.com/question/Q145682.png)
$$\int_{\mathbb{R}} \mathrm{cx}^{\mathrm{3}} \mathrm{dx}=\mathrm{1}\:\:\Rightarrow\:\mathrm{c}\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{x}^{\mathrm{3}} \mathrm{dx}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{c}\left[\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{4}} =\mathrm{1}\Rightarrow\:\mathrm{64c}=\mathrm{1} \\ $$$$\mathrm{c}=\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$\mathrm{P}\left(\mathrm{1}<\mathrm{x}<\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} \mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{256}}\left(\mathrm{16}−\mathrm{1}\right) \\ $$$$\mathrm{P}\left(\mathrm{1}<\mathrm{x}<\mathrm{2}\right)=\frac{\mathrm{15}}{\mathrm{256}}\:=\:\mathrm{0},\mathrm{059} \\ $$$$\mathrm{P}\left(\mathrm{1}<\mathrm{x}<\mathrm{2}\right)=\:\mathrm{5},\mathrm{9\%}.. \\ $$
Answered by Dwaipayan Shikari last updated on 07/Jul/21
$${f}\left({x}\right)={cx}^{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} {cx}^{\mathrm{3}} {dx}=\mathrm{1}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{64}} \\ $$$${P}\left(\mathrm{1}<{x}<\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} {dx}=\frac{\mathrm{1}}{\mathrm{256}}\left(\mathrm{2}^{\mathrm{4}} −\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{256}} \\ $$