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?

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sonu36 sonu36 · Answer 1 · 2019-11-13T09:35:37+01:00

Answer:

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

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

Step-by-step explanation:

For the banner-tailed Kangaroo rats, X has an exponential distribution with parameter [tex]\lambda[/tex] = 0.01386

So, probability distribution of X is given by,

[tex]f_{X}(x)[/tex] = [tex]\lambda \times {e^{-(\lambda \times x)}}[/tex]

for 0 ≤ x < ∞ where [tex]\lambda[/tex] = 0.01386

= 0 otherwise

so,

P( X ≤ 100) =[tex]\int_{0}^{100}(\lambda \times {e^{-(\lambda \times x)}})dx[/tex]

= [tex][- e^{-x \times \lambda}]_{0}^{100}[/tex]----------------(2)

=1 - [tex]e^{- 100 \times 0.01386}}[/tex]

= 0.7499263989

so , P(X ≤ 200)

= 1 - [tex]e^{- 200 \times 0.01386}}[/tex]

= 0.937463194

=