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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 is the probability that 6 or more were male? What is the probability that 6 on more were female? What is the probability that fewer than 4 were female? (b) Answer part (a) for the period from 1918 to the present.

Short Answer

Expert verified
Calculate the required probabilities using the binomial distribution method for both time periods, focusing on probabilities for males and females as described.

Step by step solution

01

Understanding Probability in Binomial Distribution

We have a binomial distribution problem where each wolf is either male or female. Our task is to calculate probabilities for different scenarios using the binomial probability formula. The formula for the probability of exactly k successes in n trials is given by: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n \) is the number of trials, \( k \) is the number of successes, \( p \) is the success probability on an individual trial, and \( \binom{n}{k} \) is the binomial coefficient.
02

Calculate Probabilities Before 1918

Given: 55% male before 1918 (so \( p = 0.55 \)) and 45% female (so \( q = 0.45 \)).- For \( 6 \) or more males: Calculate \( P(X \geq 6) \) where \( n = 12 \) and \( p = 0.55 \). This includes summing probabilities from 6 to 12 males.- For \( 6 \) or more females equivalent to \( (12-6) \) or more females: Calculate \( P(Y \geq 6) \) where \( Y = 12 - X \) and \( q = 0.45 \).- For fewer than 4 females, we calculate \( P(Y < 4) \) directly.
03

Calculate Probabilities for 1918 to Present

Given: 70% male from 1918 onward (so \( p = 0.70 \)) and 30% female (so \( q = 0.30 \)).- For \( 6 \) or more males: Calculate \( P(X \geq 6) \) using \( n = 12 \) and \( p = 0.70 \).- For \( 6 \) or more females, calculate \( P(Y \geq 6) \) using \( Y = 12 - X \) and \( q = 0.30 \).- For fewer than 4 females, calculate \( P(Y < 4) \) with the same probability.
04

Use Complement Rule and Cumulative Probability Tables

When calculating cumulative probabilities, it is often easier to use \( 1 - P(X < 6) \) to find \( P(X \geq 6) \). Similarly, use \( 1 - P(Y < 6) \) for \( P(Y \geq 6) \). Cumulative probability tables or a calculator can simplify these calculations. This must be done separately before 1918 and from 1918 onward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a measure of how likely an event is to occur. In the context of our wolf population problem, it involves calculating the likelihood of certain numbers of male or female wolves being observed in a sample. Probability ranges from 0, indicating impossibility, to 1, indicating certainty. For example, if there's a 55% chance a wolf spotted before 1918 in the New Mexico and Arizona region was male, we express this probability as 0.55. This means that for any single wolf observed, there is a 55% chance it will be male based on historical data.

Understanding probability helps us predict outcomes within a defined set of parameters. For example, in a random sample of 12 wolves, calculating the probability that 6 or more are male leverages the knowledge of historical male likelihood, using a specific statistical framework called the binomial distribution.
Random Sample
A random sample refers to a subset of individuals chosen from a larger population, entirely at random, to ensure each individual has an equal chance of selection. This method is important because it reduces bias and better represents the population being studied.

In the exercise, a random sample of 12 wolves is considered to make inferences about wolf genders before and after 1918. By randomly selecting, or ostensibly spotting, 12 wolves, we assume these wolves reflect the broader gender distribution of the wolves in the given time periods.

Random sampling is essential in statistical analysis because it provides a way to generalize findings from a small piece of data to a larger population, assuming that the sample is representative.
Success Trials
In binomial distribution problems, success is defined as the outcome of interest. In our wolf example, depending on the question, 'success' might be spotting a male or female wolf in a sample.

Each observation or trial has only two possible outcomes – success or failure. If we define spotting a male wolf as a success, then the probability of success (p) is the proportion of male wolves.

For instance, before 1918, where 55% are male ( p = 0.55 ), a 'success' trial would mean identifying a male wolf. Similarly, after 1918, with 70% male wolves, the probability of a single 'success' trial increases to 0.70. Therefore, success trials help quantify how often we expect to see a particular outcome and are key in applying the binomial formula.
Binomial Coefficient
The binomial coefficient is a mathematical term denoted as \( \binom{n}{k} \) , representing the number of ways to choose k successes in n trials, irrespective of order. It’s a crucial part of the binomial probability formula.

In our wolf example, if we want to know the probability of exactly 6 male wolves spotted out of 12 before 1918, we apply the binomial coefficient to determine the different ways these 6 wolves could be arranged out of 12.

The formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) calculates these combinations where n is the total number of trials (wolves), and k is the number of successes (male wolves). Using this coefficient, combined with the probability of success and failure on a single trial, it becomes possible to calculate complex probabilities in binomial settings.

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Most popular questions from this chapter

The owners of a motel in Florida have noticed that in the long run, about \(40 \%\) of the people who stop and inquire about a room for the night actually rent a room. (a) Quota Problem How many inquiries must the owner answer to be \(99 \%\) sure of renting at least one room? (b) If 25 separate inquiries are made about rooms, what is the expected number of inquiries that will result in room rentals?

The Wall Street Journal reported that approximately \(25 \%\) of the people who are told a product is improved will believe that it is, in fact, improved. The remaining \(75 \%\) believe that this is just hype (the same old thing with no real improvement). Suppose a marketing study consists of a random sample of eight people who are given a sales talk about a new, improved product. (a) Make a histogram showing the probability that \(r=0\) to 8 people believe the product is, in fact, improved. (b) Compute the mean and standard deviation of this probability distribution. (c) Quota Problem How many people are needed in the marketing study to be \(99 \%\) sure that at least one person believes the product to be improved? (Hint: Note that \(P(r \geq 1)=0.99\) is equivalent to \(1-P(0)=0.99\), or \(P(0)=0.01 .)\)

Sociologists say that \(90 \%\) of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). (See the source in Problem 12.) Suppose that six married women are having coffee together one morning. What is the probability that (a) all of them dislike their mother-in-law? (b) none of them dislike their mother-in-law? (c) at least four of them dislike their mother-in-law? (d) no more than three of them dislike their mother-in-law?

Tailgating Do you tailgate the car in front of you? About \(35 \%\) of all drivers will tailgate before passing, thinking they can make the car in front of them go faster (Source: Bernice Kanner, Are You Normal?, St. Martin's Press). Suppose that you are driving a considerable distance on a two-lane highway and are passed by 12 vehicles. (a) Let \(r\) be the number of vehicles that tailgate before passing. Make a histogram showing the probability distribution of \(r\) for \(r=0\) through \(r=12\). (b) Compute the expected number of vehicles out of 12 that will tailgate. (c) Compute the standard deviation of this distribution.

Caribbean Cruise The college student senate is sponsoring a spring break Caribbean cruise raffle. The proceeds are to be donated to the Samaritan Center for the Homeless. A local travel agency donated the cruise, valued at \(\$ 2000 .\) The students sold 2852 raffle tickets at \(\$ 5\) per ticket. (a) Kevin bought six tickets. What is the probability that Kevin will win the spring break cruise to the Caribbean? What is the probability that Kevin will not win the cruise? (b) Expected earnings can be found by multiplying the value of the cruise by the probability that Kevin will win. What are Kevin's expected earnings? Is this more or less than the amount Kevin paid for the six tickets? How much did Kevin effectively contribute to the Samaritan Center for the Homeless?

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