Before 1918, approximately 55% of the wolves in a region were male, and 45% were female....

Before 1918, approximately 55% of the wolves in a region were male, and 45% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 70% of wolves in the region are male, and 30% 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.) (a) Before 1918, 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? (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?

Solutions

Expert Solution

(a) Before 1918, in a random sample of 11 wolves spotted in the region, what is the probability that 8 or more were male?

Probability, P(X >= 8) = P(X = 8) + P(X = 9) + P(X =10) + P(X = 11)

= 0.1911
=1 - BINOM.DIST(7,11,0.55,TRUE)

What is the probability that 8 or more were female?

Probability, P(X >= 8) = P(X = 8) + P(X = 9) + P(X =10) + P(X = 11)

= 0.0610

=1 - BINOM.DIST(7,11,0.45,TRUE)

What is the probability that fewer than 5 were female?

P(X<5) = 0.3971

= BINOM.DIST(4,11,0.45,TRUE)

(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?

Probability, P(X >= 8) = P(X = 8) + P(X = 9) + P(X =10) + P(X = 11)

= 0.5696

=1 - BINOM.DIST(7,11,0.7,TRUE)

What is the probability that 8 or more were female?

Probability, P(X >= 8) = P(X = 8) + P(X = 9) + P(X =10) + P(X = 11)

= 0.0043

=1 - BINOM.DIST(7,11,0.3,TRUE)

What is the probability that fewer than 5 were female?

P(X<5) = 0.7897

= BINOM.DIST(4,11,0.3,TRUE)

orchestra answered 1 year ago

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