Let X denote the distance (m) that an animal moves from Its birth site to the fi
ID: 3222360 • 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 lambda = 0.01347. 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.) at most 100 m ________ at most 200 m ________ between 100 and 200 m What is the probability that distance exceeds the mean distance by more than 2 standard deviations? (Round your answer to four decimal places.) _______ What is the value of the median distance? _______ m Only 5% of animals will move farther than what distance? __________ mExplanation / Answer
Solution:
Given = 0.01347
mean = 1/ = 74.24
standard deviation = 1/= 74.24
CDF, C(x) = 1 - e^(-x)
a) probability that the distance is at most 100 m
P(X100)
= 1 - e^(-0.01347*100)
= 0.7400
probability that the distance is at most 200 m
P(X 200)
= 1 - e^(-0.01347*200)
= 0.9324
probability that the distance is between 100 and 200 m
= 0.9324 - 0.7400
= 0.1924
b) P(X> mean + 2*standard deviation)
= P(X>222.72)
= 1 - P(X222.72)
= 1 - (1 - e^(-0.01347*222.72))
= 0.0498
c) The median distance is ln(2) / = ln(2) / 0.01347 = 51.45 meters
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