X=reaction time, Y=length of life of electronic components Continuous joint probability distribution:
Compute: P(0 < X < 0.6 and 0.25 < Y < 0.5)
f(x,y)={4xy,0<x<1.0,0<y<10,elsewheref(x,y)=\left\{\begin{matrix} 4xy, 0<x<1.0, 0<y<1 & \\ 0, elsewhere & \end{matrix}\right.f(x,y)={4xy,0<x<1.0,0<y<10,elsewhere
P=∫∫F(x,y)dxdyP= \int \int F(x,y) dx dyP=∫∫F(x,y)dxdy
P(0<x<0.5 + 0.25<y<0.5) fall in this region.
P=∫00.6∫0.250.54xydydxP=∫00.6[4x(y22)]0.250.5dx=∫00.62x[(12)2−(14)2]dx=(38)∫00.6xdx=38(x22)00.6=38(0.6)2=0.0675P = \int^{0.6}_0 \int^{0.5}_{0.25} 4 xy dydx \\ P = \int^{0.6}_0 [4x (\frac{y^2}{2})]^{0.5}_{0.25} dx \\ = \int^{0.6}_0 2x [(\frac{1}{2})^2 -(\frac{1}{4})^2]dx \\ = (\frac{3}{8})\int^{0.6}_0 xdx \\ = \frac{3}{8}(\frac{x^2}{2})^{0.6}_0 \\ = \frac{3}{8}(0.6)^2 \\ = 0.0675P=∫00.6∫0.250.54xydydxP=∫00.6[4x(2y2)]0.250.5dx=∫00.62x[(21)2−(41)2]dx=(83)∫00.6xdx=83(2x2)00.6=83(0.6)2=0.0675
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