Question

A wildlife conservation group is designing a monitoring study of wallaby behaviour in a remote Queensland national park. The group has decided to study several regions in the park, the boundary of which form squares with side lengths W km and areas X km². A statistician has decided to choose the regions such that the region area, X, is a uniformly distributed random variable on the interval 1 < x < a such that X - U (1, a). The statistician has deduced that W= VX is a random variable that describes the side lengths of the regions. The statistician has also deduced that w has the cumulative distribution function Fw(w) = ? (W2 – 1) . Here, the value of b and the range of W depends on a. b = 2 (a) Show that b = a-1 (b) The group choose the maximum allowable region area, a, such that the average region area is equal to 5 km? What is the average region side length, E(W)? (c) The monthly monitoring cost comprises a base rate of $500 plus $50 per km². i. Write an expression for the monitoring cost, C, in terms of the region area, X. ii. Find the average monitoring cost. iii. Find the variance of the monitoring cost.

Answer 1

(a) We know that W = V X, where X is uniformly distributed on the interval (1, a). Therefore, the probability density function of X is f(x) = 1/(a-1) for 1 < x < a.

The cumulative distribution function of W is:

Fw(w) = P(W ≤ w)

= P(V X ≤ w)

= P(X ≤ w/V)

= ∫[1, w/V] f(x) dx

= (w/V - 1)/(a - 1)

Since Fw(w) = (W2 - 1)/(b - 1), we have:

(W2 - 1)/(b - 1) = (w/V - 1)/(a - 1)

Simplifying, we get:

b = a - 1

(b) We know that the average region area is 5 km². Therefore, we have:

E(X) = (1/2)(a + 1) = 5

Solving for a, we get:

a = 9

The probability density function of X is:

f(x) = 1/8 for 1 < x < 9

The cumulative distribution function of X is:

Fx(x) = (x - 1)/8

Using the relationship between W and X, we have:

W = V X

The cumulative distribution function of W is:

Fw(w) = P(W ≤ w)

= P(V X ≤ w)

= P(X ≤ w/V)

= Fx(w/V)

Substituting a = 9, we have:

Fw(w) = (w/3 - 1)/8

The probability density function of W is:

f(w) = dFw(w)/dw

= 1/24

Therefore, the average region side length is:

E(W) = ∫[0, 9] w f(w) dw

= 9/2

(c) (i) The monitoring cost, C, is given by:

C = 500 + 50 X

Substituting W = V X, we have:

C = 500 + 50 V X

(ii) The average monitoring cost is:

E(C) = E(500 + 50 V X)

= 500 + 50 E(V X)

= 500 + 50 E(V) E(X)

We know that E(X) = 5 and E(V) =