Mathematics, 09.08.2019 20:10 judyd452
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
Answers: 3
is the shape parameter and 
is the scale parameter. Then the PDF is
![E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx](/tpl/images/0173/8191/8cab0.png)
![E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x](/tpl/images/0173/8191/044ef.png)
![E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx](/tpl/images/0173/8191/7c0a9.png)
, so that 
:![E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy](/tpl/images/0173/8191/e30f7.png)
![\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}](/tpl/images/0173/8191/61e2d.png)
![\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2](/tpl/images/0173/8191/2c753.png)
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx](/tpl/images/0173/8191/ef1c8.png)
![E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx](/tpl/images/0173/8191/6b68e.png)
![E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx](/tpl/images/0173/8191/be183.png)
![E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9](/tpl/images/0173/8191/fc374.png)
![\mathrm{Var}[X]=9-E[X]^2](/tpl/images/0173/8191/5e592.png)
![\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}](/tpl/images/0173/8191/906d6.png)
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