Question

Q6. The lifetimes of two components in a machine have the following joint pdf: f(x, y).00-x y) for 0<50-y < 50 and zero elsewhere a. What is the probability that both components are functioning 20 months from now. b. What is the probability the component with life time X would fail 3 months before the other one? c. Compute the covariance of X, Y d. Compute the expected life of the machine e. What is probability that the two components fail within three months apart?

Answer 1

Solution : ( a )

$\mathrm{P(X>20,\:\:Y>20)}$

$=\int_{20}^{30}\int_{20}^{50-x}\frac{6}{125000}(50-x-y)dydx\quad\:-----\rightarrow (1)$

$\mathrm{From\:\:(1);}$

$\int _{20}^{50-x}\frac{6}{125000}\left(50-x-y\right)dy$

$\mathrm{Compute\:the\:indefinite\:integral}:\quad \int \frac{6}{125000}\left(50-x-y\right)dy$

$\int \frac{6}{125000}\left(50-x-y\right)dy$

$\mathrm{Take\:the\:constant\:out}:\quad \int a* f\left(x\right)dx=a*\int f\left(x\right)dx$

$=\frac{6}{125000}* \int \:(50-x-y)dy$

$=\frac{3}{62500}* \int \:(50-x-y)dy$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\frac{3}{62500}\left(\int \:50dy-\int \:xdy-\int \:ydy\right)$

$=\frac{3}{62500}\left(50y-xy-\frac{y^2}{2}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=\frac{3}{62500}\left(50y-xy-\frac{y^2}{2}\right)&plus;C$

$\mathrm{Compute\:the\:boundaries}:\quad \int _{20}^{50-x}\frac{6}{125000}\left(50-x-y\right)dy$

$\int _{20}^{50-x}\frac{6}{125000}\left(50-x-y\right)dy=\lim _{y\to \:50-x-}\left(\frac{3}{62500}\left(50y-xy-\frac{y^2}{2}\right)\right)-\lim _{y\to \:20&plus;}\left(\frac{3}{62500}\left(50y-xy-\frac{y^2}{2}\right)\right)$

$=\frac{3}{62500}\left(50\left(50-x\right)-x\left(50-x\right)-\frac{\left(50-x\right)^2}{2}\right)-\frac{3}{62500}\left(50*20-x*20-\frac{20^2}{2}\right)$

$=\frac{3\left(50\left(50-x\right)-x\left(50-x\right)-\frac{\left(50-x\right)^2}{2}\right)}{62500}-\frac{3\left(50*20-x*20-\frac{(20)^2}{2}\right)}{62500}$

$=\frac{\frac{3\left(x^2-100x&plus;2500\right)}{2}}{62500}-\frac{3\left(1000-20x-\frac{400}{2}\right)}{62500}$

$=\frac{3\left(x^2-100x&plus;2500\right)}{2*62500}-\frac{3\left(-20x&plus;1000-200\right)}{62500}$

$=\frac{3\left(x^2-100x&plus;2500\right)}{125000}-\frac{3\left(-20x&plus;800\right)}{62500}$

$=\frac{3\left(x^2-100x&plus;2500\right)}{125000}-\frac{3\left(-x&plus;40\right)}{3125}$

$=\frac{3}{125000}\left(x^2-100x&plus;2500\right)-\frac{3}{3125}\left(-x&plus;40\right)$

$\mathrm{Simplify}$

$=\frac{27}{1250}&plus;\frac{3x^2-180x}{125000}$

$\mathrm{So,\quad\:}\int _{20}^{50-x}\frac{6}{125000}\left(50-x-y\right)dy=\frac{27}{1250}&plus;\frac{3x^2-180x}{125000}\quad\:-\rightarrow (2)$

$\mathrm{Substitute\:(2)\:in\:(1);}$

$=\int _{20}^{30}\left(\frac{27}{1250}&plus;\frac{3x^2-180x}{125000}\right)dx$

$\int _{20}^{30}\left ( \frac{27}{1250}&plus;\frac{3x^2-180x}{125000} \right )dx$

$\mathrm{Compute\:the\:indefinite\:integral}:\quad \int \left ( \frac{27}{1250}&plus;\frac{3x^2-180x}{125000} \right )dx$

$\int \left ( \frac{27}{1250}&plus;\frac{3x^2-180x}{125000} \right )dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int \frac{27}{1250}dx&plus;\int \frac{3x^2-180x}{125000}dx$

$=\frac{27}{1250}x&plus;\frac{1}{125000}* \int \:\left ( 3x^2-180x \right )dx$

$=\frac{27}{1250}x&plus;\frac{1}{125000}\left(\int \:3x^2dx-\int \:180xdx\right)$

$=\frac{27}{1250}x&plus;\frac{1}{125000}\left(x^3-90x^2\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=\frac{27}{1250}x&plus;\frac{1}{125000}\left(x^3-90x^2\right)&plus;C$

$\mathrm{Compute\:the\:boundaries}:\quad \int _{20}^{30}\left(\frac{27}{1250}&plus;\frac{3x^2-180x}{125000}\right)dx$

$\int _{20}^{30}\left(\frac{27}{1250}&plus;\frac{3x^2-180x}{125000}\right)dx=\lim _{x\to \:30-}\left(\frac{27}{1250}x&plus;\frac{1}{125000}\left(x^3-90x^2\right)\right)-\lim _{x\to \:20&plus;}\left(\frac{27}{1250}x&plus;\frac{1}{125000}\left(x^3-90x^2\right)\right)$

$=\left ( \frac{27}{1250}*30&plus;\frac{1}{125000}\left((30)^3-90*(30)^2\right) \right )-\left ( \frac{27}{1250}*20&plus;\frac{1}{125000}\left((20)^3-90*(20)^2\right) \right )$

$=\left ( \frac{81}{125}-\frac{54}{125} \right )-\left ( \frac{54}{125}-\frac{28}{125} \right )$

$=\left ( \frac{81-54}{125} \right )-\left ( \frac{54-28}{125} \right )$

$=\frac{27}{125}-\frac{26}{125}$

$\mathrm{Simplify}$

$=\frac{1}{125}$

$=0.008$

$\mathrm{Hence,\quad\:}\int _{20}^{30}\int _{20}^{50-x}\frac{6}{125000}\left(50-x-y\right)dydx=\frac{1}{125}\quad \left(\mathrm{Decimal:\quad }\:0.008\right)$