Question

1. Before 1918, approximately 60% of the wolves in a 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 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 10 wolves spotted in the region, what is the probability that 7 or more were male?

What is the probability that 7 or more were female?

What is the probability that fewer than 4 were female?

(b) For the period from 1918 to the present, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male?

What is the probability that 7 or more were female?

What is the probability that fewer than 4 were female?

2. In a binomial situation, p + q = 1.00.

True

False    

Answer 1

(a)

Before 1918,

probability of a female wolves = 0.4

probability of a male wolves = 0.6

Let X be the number of female wolves in random sample of 10. Then, X ~ Binomial(n = 10, p = 0.4)

Probability that 7 or more were female = P(X 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= ¹⁰C₇ 0.4⁷ * 0.6³ + ¹⁰C₈ 0.4⁸ * 0.6² + ¹⁰C₉ 0.4⁹ * 0.6¹ + ¹⁰C₁₀ 0.4¹⁰ * 0.6⁰

= 0.05476188

Probability that fewer than 4 were female = P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= ¹⁰C₀ 0.4⁰ * 0.6¹⁰ + ¹⁰C₁ 0.4¹ * 0.6⁹ + ¹⁰C₂ 0.4² * 0.6⁸ + ¹⁰C₃ 0.4³ * 0.6⁷

= 0.3822806

(b)

For the period from 1918 to the present

probability of a female wolves = 0.3

probability of a male wolves = 0.7

Let X be the number of female wolves in random sample of 10. Then, X ~ Binomial(n = 10, p = 0.3)

Let Y be the number of male wolves in random sample of 10. Then, Y ~ Binomial(n = 10, p = 0.7)

Probability that 7 or more were male = P(X 7) = P(Y = 7) + P(Y = 8) + P(Y = 9) + P(Y = 10)

= ¹⁰C₇ 0.7⁷ * 0.3³ + ¹⁰C₈ 0.7⁸ * 0.3² + ¹⁰C₉ 0.7⁹ * 0.3¹ + ¹⁰C₁₀ 0.7¹⁰ * 0.3⁰

= 0.6496107

Probability that 7 or more were female = P(X 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= ¹⁰C₇ 0.3⁷ * 0.7³ + ¹⁰C₈ 0.3⁸ * 0.7² + ¹⁰C₉ 0.3⁹ * 0.7¹ + ¹⁰C₁₀ 0.3¹⁰ * 0.7⁰

= 0.01059208

Probability that fewer than 4 were female = P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= ¹⁰C₀ 0.3⁰ * 0.7¹⁰ + ¹⁰C₁ 0.3¹ * 0.7⁹ + ¹⁰C₂ 0.3² * 0.7⁸ + ¹⁰C₃ 0.3³ * 0.7⁷

= 0.6496107

2.

It is True that in a binomial situation, p + q = 1.00