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?
Accepted Solution
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Answer:P( The distance is at most 100 m) = 0.7499263989P( The distance is at most 200 m) = 0.937463194Step-by-step explanation:For the banner-tailed Kangaroo rats, X has an exponential distribution with parameter [tex]\lambda[/tex] = 0.01386So, 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 otherwiseso, 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] = 0.7499263989so , P(X ≤ 200) = 1 - [tex]e^{- 200 \times 0.01386}}[/tex] = 0.937463194 =