Question

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?

Answer 1

Answer:

the expected area of the resulting circular region is 616.38 m²

Step-by-step explanation:

Given that:

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

The expected area of the resulting circular region  is:

= $E(\pi r^2)$

= $\pi E (r^2)$

To calculate $E(r^2)$

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

$E(r^2) = \int\limits^{15}_{13} \ \dfrac{3r^2}{4}(1-(14-r)^2)dr$

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

$E(r^2) = \dfrac{3}{4} \int\limits^{15}_{13} \ r^2 (28r^3-r^4-195r^2)dr$

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

$E(r^2) = \dfrac{3}{4} [ \dfrac{28 \times 50625}{4} - \dfrac{759375}{5} - \dfrac{195 \times 3375}{3} ]-[ \dfrac{28 \times 28561}{4} - \dfrac{371293}{5} - \dfrac{195 \times 2197}{3} ]$

$E(r^2) = \dfrac{3}{4} [ 354375-151875-219375-199927+74258.6+142805]$

$E(r^2) = \dfrac{3}{4} [261.6]$

$E(r^2) = 196.2$

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:

= $\pi \times 196.2$

= 616.38 m²