Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for bann

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 λ = 0.01422.
(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.)(c) What is the value of the median distance? (Round your answer to two decimal places.)

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Parker 2 weeks 2021-09-13T13:48:38+00:00 1 Answer 0

Answers ( )

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    2021-09-13T13:49:53+00:00

    Answer:

    a) P(X<100)=0.7855

    P(X<200)=0.9418

    P(100<X<200)=0.1830

    b) P(X>M+2σ)=0.0498

    c) Median = 48.74 m

    Step-by-step explanation:

    If the distance X is modeled by an exponential distribution, with parameter λ=0.01422, we have:

    The cumulative probability function can be expressed as:

    F(x)=1-e^{-\lambda x}=1-e^{-0.01422 x}

    a) The probability that the distance is at most 100 m is:

    F(100)=1-e^{-0.01422\cdot 100}=1-e^{-1.422}=1-0.2412=0.7588

    The probability that the distance is at most 200 m is:

    F(200)=1-e^{-0.01422\cdot 200}=1-e^{-2.844}=1-0.0582=0.9418

    The probability that the distance is between 100 and 200 m is:

    F(200)-F(100)=0.9418-0.7588=0.1830

    b) The mean is M=1/λ=70.3235 and the standard deviation σ=1/λ=70.3235.

    So we have to calculate the probability that X exceeds 2 times the s.d. from the mean:

    x=M+2\sigma=70.3235+2*70.3235=210.97\\\\P(x>210.97)=1-F(210.97)\\\\P(x>210.97)=e^{-0.01422*210.97}=e^{3}=0.0498

    c) The value of the median distance is

    Median=ln(2)/\lambda=0.69314/0.01422=48.74

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