Question

Two methods of measuring surface smoothness are used to evaluate a paper product. The measurements are recorded as deviations from the nominal surface smoothness in coded units. The joint probability distribution of the two measurements is a uniform distribution over the region 0

Answer 1

`\boxed{c=(2)/(15)}`

Explanation:

The region R of integration is depicted on the picture attached.

To find the area under the surface f(x,y) = c on that region we must evaluate

`\iint_R \ f(x,y) \,dx\,dy=\iint_R \ c \,dx\,dy=c\iint_R \  \,dx\,dy`

But, by watching the region R we can see

`c\iint_R \  \,dx\,dy=c\left ( \int_(0)^(1)\int_(0)^(x+1)dxdy+\int_(1)^(4)\int_(x-1)^(x+1)dxdy \right )`

Evaluating the integrals

`\int_(0)^(1)\int_(0)^(x+1)dxdy=\int_(0)^(1)\left ( \int_(0)^(x+1) dy\right )dx=\int_(0)^(1)(x+1)dx=\\\ \int_(0)^(1)xdx+\int_(0)^(1)dx=(3)/(2)`

`\int_(1)^(4)\int_(x-1)^(x+1)dxdy \right=\int_(1)^(4)\left ( \int_(x-1)^(x+1)dy \right )dx=\int_(1)^(4)2dx=6`

And we have that the area under the surface is

`c((3)/(2)+6)=(15)/(2)c`

Since f is a probability density function

`(15)/(2)c=1`

and we must have

`\boxed{c=(2)/(15)}`