Before 1918, approximately 60% of the wolves in the New Mexico and Arizona regio
ID: 2936116 • Letter: B
Question
Before 1918, approximately 60% of the wolves in the New Mexico and Arizona region were male, and 40% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 65% of wolves in the region are male, and 35% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (Round your answers to three decimal places.)
What is the probability that 8 or more were female?
What is the probability that fewer than 5 were female?
(b) For the period from 1918 to the present, in a random sample of 11 wolves spotted in the region, what is the probability that 8 or more were male?
What is the probability that 8 or more were female?
What is the probability that fewer than 5 were female?
Explanation / Answer
Probability that a randomly selected wolf be a male, p = 0.65
Probability that 8 or more were male
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)
Using binomial distribution,
P(X = 8) = 11C8*p^8*(1-p)^3 = 0.2254
P(X = 9) = 11C9*p^9*(1-p)^2 = 0.1396
P(X = 10) = 11C10*p^10*(1-p)^1 = 0.0518
P(X = 11) = 11C11*p^11*(1-p)^0 = 0.0088
P(X >= 8) = 0.4256
Probability that a randomly selected wolf be a female, p = 0.35
Probability that 8 or more were female
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)
Using binomial distribution,
P(X = 8) = 11C8*p^8*(1-p)^3 = 0.0102
P(X = 9) = 11C9*p^9*(1-p)^2 = 0.0018
P(X = 10) = 11C10*p^10*(1-p)^1 = 0.000197
P(X = 11) = 11C11*p^11*(1-p)^0 = 0.000010
P(X >= 8) = 0.01224
Probability that fewer than were female
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
Using binomial distribution,
P(X = 0) = 11C0*p^0*(1-p)^11 = 0.008750783
P(X = 1) = 11C1*p^1*(1-p)^10 = 0.051831562
P(X = 2) = 11C2*p^2*(1-p)^9 = 0.139546513
P(X = 3) = 11C3*p^3*(1-p)^8 = 0.225421
P(X = 4) = 11C4*p^4*(1-p)^7 = 0.242761
P(X < 5) = 0.6683
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