Question #277141

Let X denote the diameter of an armores electric cable and Y denote the diameter of the ceramic mold that makes the cable. Both X and Y are scaled so that they range between 0 and 1. Suppose that X and Y have the joint density f(x,y) ={ 1/y, 0<x<y<1, 0, elsewhere

1
Expert's answer
2021-12-09T03:22:38-0500

We need the probability,

p(X+Y>12)=1p(X+Y<12)p(X+Y\gt{1\over2})=1-p(X+Y\lt{1\over2})

We have to find the limits of integration before finding this probability as follows,

Let X+Y=12    Y=12xX+Y={1\over2}\implies Y={1\over2}-x. The upper limit for Y is (1/2-x)

The upper limit for X can be determined by finding the point of intersection between the lines,

y=x...(1)y=x...(1)

and

y=12x...(2)y={1\over 2}-x...(2)

Solving for x by substituting for y in equation (2) we have,

x=12x    x=14x={1\over2}-x\implies x={1\over 4}

The lower limits for X and Y are 0 and x respectively.

Now,

1p(X+Y<12)=1014x12xf(x,y)dydx1-p(X+Y\lt{1\over2})=1-\displaystyle\int^{1\over4}_0\displaystyle\int^{{1\over2}-x}_xf(x,y)dydx


=1014x12x(1y)dydx=1-\displaystyle\int^{1\over4}_0\displaystyle\int^{{1\over2}-x}_x ({1\over y})dydx


=1014[ln(12x)lnx]dx=1-\displaystyle\int^{1\over 4}_0\lbrack ln({1\over 2}-x)-lnx\rbrack dx


=1+[(12x)ln(12x)xlnx]014=1+\lbrack({1\over 2}-x)ln({1\over 2}-x)-xlnx\rbrack|^{1\over4}_0


=1+12ln(12)=0.6534=1+{1\over 2}ln({1\over2})=0.6534


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