The average squared distance between the points is their variance
The average squared distance is 8/3
<h3>How to determine the average squared distance?</h3>
The given parameters are:
R={(x,y): 0<=x<=2, 0<=y<=2}
The point = (2,2).
f(x,y) = (x-2)² + (y-2)²
The squared distance is calculated as:
D² = f(x,y) = (x-2)² + (y-2)²
Where the area (A) is:
A = xy
Substitute the maximum values of x and y in A = xy
A = 2 * 2
A = 4
The minimum values of x and y are 0.
So, the limits for the integrals are 0 to 2
The integral becomes

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Evaluate the inner integral with respect to x from 0 to 2

Expand
![D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy](https://tex.z-dn.net/?f=D%5C%C2%B2%20%3D%20%5Cint%20%5B%5Cfrac%2013%282%29%5E3%20-2%282%29%5E2%20%2B%204%282%29%20%2B%20%282%29%28y-2%29%5C%C2%B2%5D%20dy)
Simplify

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Evaluate the integral with respect to y from 0 to 2
![D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0](https://tex.z-dn.net/?f=D%5C%C2%B2%20%3D%20%5B%5Cfrac%2083y%20%2B%20%5Cfrac%2023y%5E3-%204y%5E2%20%2B%208y%20%5D%7C%5Climits%5E2_0)
Expand

D² = 32/3
The average squared distance (AD²) is calculated as:
AD² = D²/A
So, we have:
AD² = 32/3
4
Evaluate the quotient
AD² = 32/12
Simplify
AD² = 8/3
Hence, the average squared distance is 8/3
Read more about average distance or variance at:
brainly.com/question/15858152