Question #177868

Statistics and Probability 10. f(x) = 2w e-2x + 3(1-we-35 with w-U(0,1). Calculate: a. EX) b. VOX ProX > 1

1
Expert's answer
2021-04-15T06:56:55-0400

f(x)=2we2x+3(1w)e3xf(x)=2we^{-2x}+3(1-w)e^{-3x}

fX(x)=01(2we2x+3(1w)e3x)dw=e2x3e3x2f_X(x)=\displaystyle\intop^1_0(2we^{-2x}+3(1-w)e^{-3x})dw=e^{-2x}-\frac{3e^{-3x}}{2}


a.

E(X)=0x(e2x3e3x2)dx=E(X)=\displaystyle\intop^{\infin}_{0}x(e^{-2x}-\frac{3e^{-3x}}{2})dx=


=((2x+1)e2x4+(3x+1)e3x6)0=112=(-\frac{(2x+1)e^{-2x}}{4}+\frac{(3x+1)e^{-3x}}{6})|^{\infin}_{0}=\frac{1}{12}


b.

V(X)=0x2(e2x3e3x2)dx(E(X))2=V(X)=\displaystyle\intop^{\infin}_{0}x^2(e^{-2x}-\frac{3e^{-3x}}{2})dx-(E(X))^2=


=((2x2+2x+1)e2x4+(9x2+6x+2)e3x18)01144=5361144=19144=(-\frac{(2x^2+2x+1)e^{-2x}}{4}+\frac{(9x^2+6x+2)e^{-3x}}{18})|^{\infin}_{0}-\frac{1}{144}=\frac{5}{36}-\frac{1}{144}=\frac{19}{144}


c.

P(X>1)=1(e2x3e3x2)dx=P(X>1)=\displaystyle\intop^{\infin}_{1}(e^{-2x}-\frac{3e^{-3x}}{2})dx=


=(e2x2+e3x2)1=e22e32=0.0428=(-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{2})|^{\infin}_1=\frac{e^{-2}}{2}-\frac{e^{-3}}{2}=0.0428


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