Question

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for banner-tailed kangaroo rats, X has an exponential distribution with parameter λ = 0.01372.

What is the probability that the distance is at most 100 m? At most 200 m? Between 100 and 200 m?

Answer 1

74.64% probability that the distance is at most 100 m

93.57% probability that the distance is at most 200 m

18.93% probability that the distance is between 100 and 200m.

Explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

`f(x) = \mu e^(-\mu x)`

In which
`\mu = (1)/(m)` is the decay parameter.

The probability that x is lower or equal to a is given by:

`P(X \leq x) = \int\limits^a_0 {f(x)} \, dx`

Which has the following solution:

`P(X \leq x) = 1 - e^(-\mu x)`

In this problem, we have that:

`\mu = 0.01372`

What is the probability that the distance is at most 100 m?

`P(X \leq 100) = 1 - e^(-0.01372*100) = 0.7464`

74.64% probability that the distance is at most 100 m

At most 200 m?

`P(X \leq 200) = 1 - e^(-0.01372*200) = 0.9357`

93.57% probability that the distance is at most 200 m

Between 100 and 200 m?

0.9357 - 0.7464 = 0.1893

18.93% probability that the distance is between 100 and 200m.