Question

(1 point) Let p be the joint density function such that p(x,y)=116xy in R, the rectangle 0≤x≤4,0≤y≤2, and p(x,y)=0 outside R. Find the fraction of the population satisfying the constraint x≥y fraction

Answer 1

Answer:

The answer is $\frac{7}{8}$ of the population.

Step-by-step explanation:

The question is wrong. The joint density function is $p(x,y)=(\frac{1}{16})xy$ in R

and $p(x,y)=0$ outside R.

R is defined as the rectangle $0\leq x\leq 4$ , $0\leq y\leq 2$

In order to find the fraction of the population satisfying the constraint $x\geq y$ , we will need to integrate the joint density function $p(x,y)$ over the region defined by the constraint. It is very convenient to draw the region ''R'' and the new region define by the constraint $x\geq y$

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 $p(x,y)=0$ outside ''R'', the integral will be equal to 0 (because of the joint density function).

Inside the rectangle ''R'' and given the constraint $x\geq y$ , 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 ⇒

$\int\int p(x,y) dx dy$ ⇒

$\int \int\limits_1 {p(x,y)} \, dx dy + \int \int\limits_2{p(x,y)} \, dx dy$ , where ''1'' is the green region and ''2'' is the blue region.

⇒ $\int\limits^2_0 \int\limits^x_0 (\frac{1}{16} xy) dy dx$   +  $\int\limits^4_2\int\limits^2_0 (\frac{1}{16}xy) dydx$  = $\frac{1}{8}+\frac{3}{4}=\frac{7}{8}=0.875$

We find that $\frac{7}{8}$ of the population satisfy the constraint $x\geq y$.