Let X denote the distance (in meters) that an animal moves from its birth site t
ID: 3384157 • Letter: L
Question
Let X denote the distance (in meters) 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 mean 72.15 meters.
(a) What is the probability that the distance is at most 100 meters? At most 200 meters? Between 100 and 200 meters?
(b) What is the probability that the distance exceeds the mean distance by more than 2 standard deviations?
(c) What is the value of the median distance?
Explanation / Answer
We have given that,
The random variable X is the distance (in meters) that an animal moves from its birth site to the first territorial vacancy it encounters.
X ~ Exp( = )
We have given that mean of the distribution is 72.15.
The parameter of exponential distribution is .
= 1 / mean = 1 / 72.15 = 0.01386
The p.d.f. of exponential distribution is,
f(X) = * e-x
F(x) = P( X x ) = 1 - e-x
Find the above probability :
(a) The distance is at most 100 meters?
P(X 100) = 1 - e-0.01386 * 100 = 1 - 0.250074 = 0.749926 which is approximately 0.75
b) The distance is at most 200 meters?
P(X 200) = 1 - e-0.01386 * 200 = 0.9375
c) Between 100 and 200 meters?
P(100 X 200) = P(X 200) - P( X 100) = 0.9375 - 0.7500 = 0.1875
d) What is the probability that the distance exceeds the mean distance by more than 2 standard deviations?
two standard deviations is two mean values added which is 72.15 + 72.15 = 144.3
P(X > 144.3) = 1 - P(X 144.3) = 1 - [ 1 - e-0.01386*144.3 ] = 1 - [ 1 - 0.1353 ] = 0.1353
e) What is the value of the median distance?
The median of the Exponential distribution is,
Median = ln(2) / = ln(2) / 0.01386 = 50.01 meters
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