Respuesta :
The solutions and probabilities are,
(a)
P(X < 100) = 0.748
P(X < 200) = 0.936
P(100 < X < 200) = 0.188
(b) P(X > M+2σ) = 0.0498
(c) Median = 50.01 m
Given data,
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 = 0.01386
[tex]F(x) = 1 - e^-^λ^x[/tex]
(a)
The probability that the distance is at most 100 m is,
[tex]F(100) = 1 - e^-^0^.^0^1^3^8^*^1^0^0[/tex] = 0.748
The probability that the distance is at most 200 m is,
[tex]F(200) = 1 - e^-^0^.^0^1^3^8^*^2^0^0[/tex] = 0.936
The probability that the distance is between 100 and 200 m is,
F(200) - F(100) = 0.936 - 0.748 = 0.188
(b)
The probability that distance exceeds the mean distance by more than 2 standard deviations
Mean, μ = 1/λ = 1/0.01386 = 72.1501
Standard deviation, σ = 1/λ = 72.1501
Value more than 2 SD from mean = µ + 2*σ
= 72.1501 + 2*72.1501 = 216.4503
Probability of greater than 216.4503,
P(X > 216.4503) = e^(-0.0139 * 216.4503)
= 0.0498
(c)
The value of the median distance is,
Median = ln(2) / λ
= ln(2) / 0.0138 = 50.01
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