The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3.† Compute the following. (Round your answers to three decimal places.) (a) E(X) and V(X) E(X) = 2.659 V(X) = 1.931 (b) P(X ≤ 3) .6321 (c) P(1 ≤ X ≤ 3) .5269

I'm assuming [tex]\alpha[/tex] is the shape parameter and [tex]\beta[/tex] is the scale parameter. Then the PDF is[tex]f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}[/tex]a. The expectation is[tex]E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx[/tex]To compute this integral, recall the definition of the Gamma function,[tex]\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt[/tex]For this particular integral, first integrate by parts, taking[tex]u=x\implies\mathrm du=\mathrm dx[/tex][tex]\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}[/tex][tex]E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x[/tex][tex]E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx[/tex]Substitute [tex]x=3y^{1/2}[/tex], so that [tex]\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy[/tex]:[tex]E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy[/tex][tex]\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}[/tex]The variance is[tex]\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2[/tex]The second moment is[tex]E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx[/tex]Integrate by parts, taking[tex]u=x^2\implies\mathrm du=2x\,\mathrm dx[/tex][tex]\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}[/tex][tex]E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx[/tex][tex]E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx[/tex]Substitute [tex]x=3y^{1/2}[/tex] again to get[tex]E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9[/tex]Then the variance is[tex]\mathrm{Var}[X]=9-E[X]^2[/tex][tex]\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}[/tex]b. The probability that [tex]X\le3[/tex] is[tex]P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx[/tex]which can be handled with the same substitution used in part (a). We get[tex]\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}[/tex]c. Same procedure as in (b). We have[tex]P(1\le X\le3)=P(X\le3)-P(X\le1)[/tex]and[tex]P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}[/tex]Then[tex]\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}[/tex]