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An ecologist wishes to mark off a circular sampling region having radius 10 m. However, the radius of the resulting region is actually a random variable R with the following pdf.

f(r)={34(1−(14−r)2)13≤r≤150 otherwise
What is the expected area of the resulting circular region?

User Hemu
by
5.5k points

1 Answer

3 votes

Answer:

the expected area of the resulting circular region is 616.38 m²

Explanation:

Given that:


f(r) = \left \{ {{(3)/(4)(1-(14-r)^2)} \atop {0 }} \right. \ \ 13 \leq r \leq 15 otherwise

The expected area of the resulting circular region is:

=
E(\pi r^2)

=
\pi E (r^2)

To calculate
E(r^2)


E(r^2) = \int\limits^(15)_(13) {r^2} \ f(r) \ dr


E(r^2) = \int\limits^(15)_(13) \ (3r^2)/(4)(1-(14-r)^2)dr


E(r^2) = (3)/(4) \int\limits^(15)_(13) \ r^2 (1-196-r^2+28r) dr


E(r^2) = (3)/(4) \int\limits^(15)_(13) \ r^2 (28r^3-r^4-195r^2)dr


E(r^2) = (3)/(4)[(28 r^4)/(4)-(r^5)/(5)-(195r^3)/(3)]^{^(15)}}__(13)


E(r^2) = (3)/(4) [ (28 * 50625)/(4) - (759375)/(5) - (195 * 3375)/(3) ]-[ (28 * 28561)/(4) - (371293)/(5) - (195 * 2197)/(3) ]


E(r^2) = (3)/(4) [ 354375-151875-219375-199927+74258.6+142805]


E(r^2) = (3)/(4) [261.6]


E(r^2) = 196.2

Recall:

The expected area of the resulting circular region is:

=
E(\pi r^2)

=
\pi E (r^2)

where;


E(r^2) = 196.2

Then

The expected area of the resulting circular region is:

=
\pi * 196.2

= 616.38 m²

User Nick Bernard
by
6.0k points