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 = .01386 (as suggested in the article "Competition and Dispersal from Multiple Nests, " Ecology, 1997: 873-883). What is the probability that the distance is at most 100 m? At most 200 m?

Answer 1

P( The distance is at most 100 m) = 0.7499263989

P( The distance is at most 200 m) = 0.937463194

Explanation:

For the banner-tailed Kangaroo rats, X has an exponential distribution with parameter
`\lambda` = 0.01386

So, probability distribution of X is given by,

`f_(X)(x)` =
`\lambda * {e^(-(\lambda * x))}`

for 0 ≤ x < ∞ where
`\lambda` = 0.01386

= 0 otherwise

so,

P( X ≤ 100) =
`\int_(0)^(100)(\lambda * {e^(-(\lambda * x))})dx`

=
`[- e^(-x * \lambda)]_(0)^(100)`----------------(2)

=1 -
`e^(- 100 * 0.01386)}`

= 0.7499263989

so , P(X ≤ 200)

= 1 -
`e^(- 200 * 0.01386)}`

= 0.937463194

=