Question Number 192377 by Spillover last updated on 16/May/23
$${The}\:{probability}\:{density}\:{function}\:{f}\left({x}\right) \\ $$$$\:{of}\:{a}\:{variable}\:{x}\:{is}\:{given}\:{by}\: \\ $$$${f}\left({x}\right)=\begin{cases}{{kx}\mathrm{sin}\:\pi{x}\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}}\\{\mathrm{0}\:{for}\:{all}\:{value}\:{of}\:{x}}\end{cases} \\ $$$${Show}\:{that}\:{k}=\pi\:\:{and}\:{deduce}\:{that}\:{mean}\: \\ $$$${and}\:{the}\:{variance}\:{of}\:{the}\:{distribution}\:{are} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\right)\:{and}\:\frac{\mathrm{2}}{\pi^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\right) \\ $$
Answered by mehdee42 last updated on 16/May/23
![∫_0 ^1 kxsinπxdx=1⇒−(k/π)xcosπx+(k/π^2 )sinπx]_0 ^1 =1 ⇒k=π ✓ E(x)=∫_0 ^1 πx^2 sinπxdx=π(−(x^2 /π)cosπx+((2x)/π^2 )sinπx+(2/π^2 )cosπx]_0 ^1 π((1/π)−(4/π^3 ))=1−(4/π^2 ) ✓ E(x^2 )=∫_0 ^1 x^3 sinπxdx =π[−(x^3 /π)cosπx+((3x^2 )/π)sinπx−((6x)/π^3 )cosπx+(6/π^4 )sinπx]_0 ^1 =1−(6/π^2 ) Var(x)=E(x^2 )−E^2 (x)=1−(6/π^2 )−(1−(4/π^2 ))^2 =(2/π^2 )(1−(8/π^2 )) ✓](https://www.tinkutara.com/question/Q192386.png)
$$\left.\int_{\mathrm{0}} ^{\mathrm{1}} {kxsin}\pi{xdx}=\mathrm{1}\Rightarrow−\frac{{k}}{\pi}{xcos}\pi{x}+\frac{{k}}{\pi^{\mathrm{2}} }{sin}\pi{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow{k}=\pi\:\checkmark \\ $$$${E}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \pi{x}^{\mathrm{2}} {sin}\pi{xdx}=\pi\left(−\frac{{x}^{\mathrm{2}} }{\pi}{cos}\pi{x}+\frac{\mathrm{2}{x}}{\pi^{\mathrm{2}} }{sin}\pi{x}+\frac{\mathrm{2}}{\pi^{\mathrm{2}} }{cos}\pi{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\pi\left(\frac{\mathrm{1}}{\pi}−\frac{\mathrm{4}}{\pi^{\mathrm{3}} }\right)=\mathrm{1}−\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\:\:\checkmark \\ $$$${E}\left({x}^{\mathrm{2}} \right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{3}} \:{sin}\pi{xdx} \\ $$$$=\pi\left[−\frac{{x}^{\mathrm{3}} }{\pi}{cos}\pi{x}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\pi}{sin}\pi{x}−\frac{\mathrm{6}{x}}{\pi^{\mathrm{3}} }{cos}\pi{x}+\frac{\mathrm{6}}{\pi^{\mathrm{4}} }{sin}\pi{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:=\mathrm{1}−\frac{\mathrm{6}}{\pi^{\mathrm{2}} } \\ $$$${Var}\left({x}\right)={E}\left({x}^{\mathrm{2}} \right)−{E}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−\frac{\mathrm{6}}{\pi^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{\pi^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\right)\:\:\checkmark \\ $$$$ \\ $$
Commented by Spillover last updated on 16/May/23
$${your}\:{right}. \\ $$