Answer to Question #347823 in Statistics and Probability for Benpraiz

Question #347823

If F(x) 1/39(3x-2)2;0<_x_<3


O: elsewhere


1. Verify that F(x) is a PDF


2. find E(x) and Var(x)


1
Expert's answer
2022-06-06T02:46:55-0400

1.


F(x)dx=03(3x2)239dx\displaystyle\int_{-\infin}^{\infin}F(x)dx=\displaystyle\int_{0}^{3}\dfrac{(3x-2)^2}{39}dx

=[(3x2)33(3)(39)]30=343+8351=1,True=[\dfrac{(3x-2)^3}{3(3)(39)}]\begin{matrix} 3\\ 0 \end{matrix}=\dfrac{343+8}{351}=1, True

2.


E(X)=F(x)xdx=03x(3x2)239dxE(X)=\displaystyle\int_{-\infin}^{\infin}F(x)xdx=\displaystyle\int_{0}^{3}\dfrac{x(3x-2)^2}{39}dx

=039x312x2+4x39dx=\displaystyle\int_{0}^{3}\dfrac{9x^3-12x^2+4x}{39}dx

=[139(9x444x3+2x2)]30=[\dfrac{1}{39}(\dfrac{9x^4}{4}-4x^3+2x^2)]\begin{matrix} 3\\ 0 \end{matrix}

=139(7294108+180)=\dfrac{1}{39}(\dfrac{729}{4}-108+18-0)

=12352=\dfrac{123}{52}



E(X2)=F(x)x2dx=03x2(3x2)239dxE(X^2)=\displaystyle\int_{-\infin}^{\infin}F(x)x^2dx=\displaystyle\int_{0}^{3}\dfrac{x^2(3x-2)^2}{39}dx

=039x412x3+4x239dx=\displaystyle\int_{0}^{3}\dfrac{9x^4-12x^3+4x^2}{39}dx

=[139(9x553x4+4x33)]30=[\dfrac{1}{39}(\dfrac{9x^5}{5}-3x^4+\dfrac{4x^3}{3})]\begin{matrix} 3\\ 0 \end{matrix}

=139(21875243+360)=\dfrac{1}{39}(\dfrac{2187}{5}-243+36-0)

=38465=\dfrac{384}{65}

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

=38465(12352)2=422713520=\dfrac{384}{65}-(\dfrac{123}{52})^2=\dfrac{4227}{13520}


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