Let X denote the distance (m) that an animal moves from its birth site to the fi

Q: Let X denote the distance (m) that an animal moves from its birth site to the fi

Solution: Given = 0.01337, mean = 1/ = 74.79 a) (P

ID: 3183174 • Letter: L

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 = 0.01337.

(a) What is the probability that the distance is at most 100 m? At most 200 m? Between 100 and 200 m? (Round your answers to four decimal places.)

(b) What is the probability that distance exceeds the mean distance by more than 2 standard deviations? (Round your answer to four decimal places.)

at most 100 m ( ) at most 200 m ( ) between 100 and 200 m

Explanation / Answer

Solution:

Given = 0.01337, mean = 1/ = 74.79

a)
(P <=100) = 1 - e^(-x) x is 100 = 1-e^(-0.01337*100)
= 0.7373

(P <=200) = 1 - e^(-0.01337*200) = 0.9310

P (100 < X < 200) = 0.9310 - 0.7373 = 0.1937

b) two standard deviations is two mean values added which is 149.58
P(X > 149.58) = e ^(-x) = e^(-0.01337*149.58)= 0.1353

c)
The median distance is ln(2) / = 0.6931/0.1353 = 51.84meters

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Let X denote the distance (m) that an animal moves from its birth site to the fi

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Let X denote the distance (m) that an animal moves from its birth site to the fi

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