Answer:
The answer is [tex]\frac{7}{8}[/tex] of the population.
Step-by-step explanation:
The question is wrong. The joint density function is [tex]p(x,y)=(\frac{1}{16})xy[/tex] in R
and [tex]p(x,y)=0[/tex] outside R.
R is defined as the rectangle [tex]0\leq x\leq 4[/tex] , [tex]0\leq y\leq 2[/tex]
In order to find the fraction of the population satisfying the constraint [tex]x\geq y[/tex] , we will need to integrate the joint density function [tex]p(x,y)[/tex] over the region defined by the constraint. It is very convenient to draw the region ''R'' and the new region define by the constraint [tex]x\geq y[/tex]
I will attach a drawing with the region ''R'' and the new region where we need to apply the integral.
If we integrate outside ''R'', given that [tex]p(x,y)=0[/tex] outside ''R'', the integral will be equal to 0 (because of the joint density function).
Inside the rectangle ''R'' and given the constraint [tex]x\geq y[/tex] , we define two new regions : the green region (I) and the blue region (II).
The final step is to integrate in (I) and in (II) and sum ⇒
[tex]\int\int p(x,y) dx dy[/tex] ⇒
[tex]\int \int\limits_1 {p(x,y)} \, dx dy + \int \int\limits_2{p(x,y)} \, dx dy[/tex] , where ''1'' is the green region and ''2'' is the blue region.
⇒ [tex]\int\limits^2_0 \int\limits^x_0 (\frac{1}{16} xy) dy dx[/tex] + [tex]\int\limits^4_2\int\limits^2_0 (\frac{1}{16}xy) dydx[/tex] = [tex]\frac{1}{8}+\frac{3}{4}=\frac{7}{8}=0.875[/tex]
We find that [tex]\frac{7}{8}[/tex] of the population satisfy the constraint [tex]x\geq y[/tex].
The fraction of the population satisfying the constraint x ≥ y fraction will be 7/8.
When the two curves intersect then they bound the region is known as the area bounded by the curve.
Let p be the joint density function such that [tex]p(x,y)=\dfrac{1}{16}xy[/tex] in R, the rectangle [tex]0\leq x\leq 4,0\leq y\leq 2[/tex], and [tex]p(x,y)=0[/tex] outside R.
The fraction of the population satisfying the constraint x ≥ y fraction will be
Then we have
[tex]\int \int p(x,y)\ dx\ dy = \int \int _1 p(x,y)\ dx\ dy + \int \int _2 p(x,y)\ dx\ dy[/tex]
Where 1 is the green region and 2 is the blue region. Then we have
[tex]\rightarrow \int_{0}^{2} \int_{0}^{x} \left ( \dfrac{1}{16}xy \right )dx\ dy + \int_{2}^{4}\int \int_{0}^{2} \left ( \dfrac{1}{16}xy \right ) \\\\\\\rightarrow \dfrac{1}{8} + \dfrac{3}{4} \\\\\\ \rightarrow \dfrac{7}{8}[/tex]
We find that 7/8 of the population satisfies the constraint y ≤ x.
The graph is shown below.
More about the area bounded by the curve link is given below.
https://brainly.com/question/24563834
