Chapter 6.2, Problem 17P

In each of the following problems, the binomial distribution will be used. Answers may vary slightly depending on whether the binomial distribution formula, the binomial distribution table, or the distribution results from a calculator or computer are used. Please answer the following questions and then complete the problem.

What makes up a trial? What is a success? What is a failure?

What are the values of n, p, and q?

Ecology: Wolves The following is based on information taken from The Wolf in the Southwest: The Making of an Endangered Species, edited by David Brown (University of Arizona Press). Before 1918. approximately 55% of the wolves in the New Mexico and Arizona 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.

(a) Before 1918, in a random sample of 12 wolves spotted in the region, what was the probability that 6 or more were male? What was the probability that 6 or more were female? What was the probability that fewer than 4 were female?

(b) Answer pan (a) for the period from 1918 to the present.

To determine

The probability that 6 or more wolves were male, 6 or more wolves were female and fewer than 4 wolves were female before 1918.

Answer to Problem 17P

Solution: The probability that 6 or more wolves were male is 0.740, 6 or more wolves were female is 0.473 and fewer than 4 wolves were female is 0.135.

Explanation of Solution

Given: The sample size n is 12.

The percentage of male wolves in New Mexico and Arizona region is 55% and the percentage of female wolves is 45% before 1918.

Calculation: The trial consists of determining the sex of wolf.

Consider success as male and failure as female.

The probability of success and failure is provided below,

$\begin{array}{l}p=0.55\\ q=0.45\end{array}$

The general formula for binomial distribution to calculate probability is as follows,

$P\left(r\right)=C_{n,r}{p}^r{q}^{n-r}$

Here $n$ is the number of trials, $r$ is a random variable that represents the number of success out of $n$ trials.

In this case, the random r is the number of male wolves in a group of 12.

The probability that 6 or more wolves were male can be obtained as:

$\begin{array}{c}P\left(r\ge 6\right)=P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)\\ =C_{12,6}{\left(0.55\right)}^6{\left(0.45\right)}^{12-6}+C_{12,7}{\left(0.55\right)}^7{\left(0.45\right)}^{12-7}+\cdots +C_{12,12}{\left(0.55\right)}^{12}{\left(0.45\right)}^{12-12}\\ =0.212+0.223+0.170+\cdots +0.001\\ =0.740\end{array}$

Six or more female wolves in a group of 12 is same as less than or equal to 6 male wolves.

Now, the probability that 6 or more were female wolves can be obtained as:

$\begin{array}{c}P\left(r\le 6\right)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)\\ =C_{12,0}{\left(0.55\right)}^0{\left(0.45\right)}^{12-0}+C_{12,1}{\left(0.55\right)}^1{\left(0.45\right)}^{12-1}\cdots +C_{12,6}{\left(0.55\right)}^6{\left(0.45\right)}^{12-6}\\ =0.000+0.001+\cdots +0.212\\ =0.473\end{array}$

The probability of fewer than 4 female wolves is same as the probability of more than 8 male wolves.

Now, the probability of fewer than 4 females can be obtained as:

$\begin{array}{c}P\left(r>8\right)=P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)\\ =C_{12,9}{\left(0.55\right)}^9{\left(0.45\right)}^{12-9}+C_{12,10}{\left(0.55\right)}^{10}{\left(0.45\right)}^{12-10}+C_{12,11}{\left(0.55\right)}^{11}{\left(0.45\right)}^{12-11}+C_{12,12}{\left(0.55\right)}^{12}{\left(0.45\right)}^{12-12}\\ =0.092+0.034+0.008+0.001\\ =0.135\end{array}$

Hence, the probability that 6 or more were male wolves is 0.740, 6 or more were female wolves is 0.473 and fewer than 4 female wolves is 0.135.

To determine

The probability that 6 or more were male wolves, 6 or more were female wolves and fewer than 4 were female wolves from 1918 to present.

Answer to Problem 17P

Solution: The probability that 6 or more were male wolves is 0.961, 6 or more were female wolves is 0.117 and fewer than 4 were female wolves is 0.493.

Explanation of Solution

Given: The sample size n is 12.

The percentage of male wolves in the region is 70% and the percentage of female wolves is 30% after 1918.

Calculation: The probability of success and failure are provided below:

$\begin{array}{l}p=0.70\\ q=0.30\end{array}$

The probability that 6 or more wolves were male can be obtained as:

$\begin{array}{c}P\left(r\ge 6\right)=P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)\\ =C_{12,6}{\left(0.70\right)}^6{\left(0.30\right)}^{12-6}+C_{12,7}{\left(0.70\right)}^7{\left(0.30\right)}^{12-7}\cdots +C_{12,12}{\left(0.70\right)}^{12}{\left(0.30\right)}^{12-12}\\ =0.079+0.158+\cdots +0.014\\ =0.961\end{array}$

Six or more female wolves in a group of 12 is same as less than or equal to 6 male wolves.

Now, the probability that 6 or more were female wolves can be obtained as:

$\begin{array}{c}P\left(r\le 6\right)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)\\ =C_{12,0}{\left(0.70\right)}^0{\left(0.30\right)}^{12-0}+C_{12,1}{\left(0.70\right)}^1{\left(0.30\right)}^{12-1}\cdots +C_{12,6}{\left(0.70\right)}^6{\left(0.30\right)}^{12-6}\\ =0.000+0.000+\cdots +0.079\\ =0.117\end{array}$

The probability of fewer than 4 female wolves is same as the probability of more than 8 male.

Now, the probability of fewer than 4 female wolves can be obtained as:

$\begin{array}{c}P\left(r>8\right)=P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)\\ =C_{12,9}{\left(0.70\right)}^9{\left(0.30\right)}^{12-9}+C_{12,10}{\left(0.70\right)}^{10}{\left(0.30\right)}^{12-10}+C_{12,11}{\left(0.70\right)}^{11}{\left(0.30\right)}^{12-11}+C_{12,12}{\left(0.70\right)}^{12}{\left(0.30\right)}^{12-12}\\ =0.240+0.168+0.071+0.014\\ =0.493\end{array}$

Hence, from 1918 to present, the probability that 6 or more were male wolves is 0.961, 6 or more were female wolves is 0.117 and fewer than 4 were female wolves is 0.493.