Question

According to the Humane Society of the United States, there are approximately 77.5 million owned dogs in the United States, and approximately \(40 \%\) of all U.S. households own at least one \(\operatorname{dog} .\). Suppose that the \(40 \%\) figure is correct and that 15 households are randomly selected for a pet ownership survey. a. What is the probability that exactly eight of the households have at least one dog? b. What is the probability that at most four of the households have at least one dog? c. What is the probability that more than 10 households have at least one dog?

Short answer

Answer: The probabilities are approximately 21.78% for exactly 8 households, 12.03% for at most 4 households, and 0.03% for more than 10 households.

Step-by-step solution

Find Variables for the Binomial Formula

For this problem, we have the following variables: n (number of trials) = 15 households p (probability of success, i.e., having at least one dog) = 0.4 q (probability of failure, i.e., not having a dog) = 1 - p = 0.6 #a. Probability That Exactly 8 Households Have at Least 1 Dog#

Apply the Binomial Formula

We want to find the probability of k = 8 successes (households with at least one dog) in n = 15 trials. The binomial probability formula is: \( P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \) Plugging in the values we have: \( P(X = 8) = \binom{15}{8} \cdot 0.4^8 \cdot 0.6^{7} \)

Compute the Binomial Coefficient

The binomial coefficient (n choose k) can be computed as follows: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) In our case: \( \binom{15}{8} = \frac{15!}{8!(15-8)!} = \frac{15!}{8!7!} \)

Calculate the Probability

Now, we can calculate the probability: \( P(X = 8) = \frac{15!}{8!7!} \cdot 0.4^8 \cdot 0.6^{7} \approx 0.2178 \) So, the probability that exactly eight of the households have at least one dog is approximately 21.78%. #b. Probability That At Most 4 Households Have at Least 1 Dog#

Calculate the Probability for X ≤ 4

We now want to find the probability that at most 4 households have a dog, which means P(X ≤ 4). This is equivalent to the sum of probabilities of X = 0, 1, 2, 3, and 4: \(P(X \leq 4) = \sum_{k=0}^{4} \binom{15}{k} \cdot 0.4^k \cdot 0.6^{(15-k)} \)

Calculate Each Probability and Sum

Now, we calculate each probability and sum them up: \(P(X \leq 4) \approx \sum_{k=0}^{4} \frac{15!}{k!(15-k)!} \cdot 0.4^k \cdot 0.6^{(15-k)} \approx 0.1203 \) So, the probability that at most four of the households have at least one dog is approximately 12.03%. #c. Probability That More Than 10 Households Have at Least 1 Dog#

Calculate the Probability for X > 10

We now want to find the probability that more than 10 households have a dog, which means P(X > 10). This is equivalent to the sum of probabilities of X = 11, 12, 13, 14, and 15: \(P(X > 10) = \sum_{k=11}^{15} \binom{15}{k} \cdot 0.4^k \cdot 0.6^{(15-k)} \)

Calculate Each Probability and Sum

Now, we calculate each probability and sum them up: \(P(X > 10) \approx \sum_{k=11}^{15} \frac{15!}{k!(15-k)!} \cdot 0.4^k \cdot 0.6^{(15-k)} \approximately 0.0003 \) So, the probability that more than 10 households have at least one dog is approximately 0.03%.

Key concepts

Binomial Distribution

The binomial distribution is a key probability distribution used in statistics to model situations where there are only two possible outcomes, such as success and failure. It is perfect for scenarios like our pet ownership survey, where each household either has a dog (success) or doesn't have a dog (failure). When you apply the binomial distribution, you want to understand the likelihood of achieving a certain number of successes across many trials, like finding out how many households out of 15 have at least one dog.

Some important features of the binomial distribution include:

• A fixed number of trials, denoted by \( n \), which in the exercise is 15 households.
• Each trial has two possible outcomes.
• The probability \( p \) of success remains constant throughout the trials — here, owning a dog is 0.4.
• The binomial probability formula \( P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \), helps compute the probability of having \( k \) successes in \( n \) trials.

This distribution provides a powerful way to understand binary scenarios, especially when outcomes align perfectly with the model assumptions.

Probability Theory

Probability theory underpins our ability to predict how likely certain events are to occur. In simple terms, it's the branch of mathematics that helps us deal with uncertainty and it provides rules for calculating drastic chances.

In our exercise, we deal with theoretical probability. This is where outcomes (like having a dog) are consistent and based on a known probability (40% of households own a dog). The probability for an exact outcome, such as exactly 8 households owning a dog in our sample, is found using the binomial probability formula. This formula relies on factorial computation, which sounds tricky, but it simply means multiplying sequences of decreasing positive integers.

Effective use of probability theory allows clear forecasting in statistics and everyday decision-making. It helps us draw meaningful conclusions from data, like determining the statistical likelihood of different pet ownership scenarios.

Statistical Analysis

Statistical analysis is the process of collecting, examining, and interpreting quantitative data to discover meaningful patterns and trends. In the exercise, we're analyzing household survey data to assess dog ownership rates across different households.

We use statistical analysis to calculate probabilities, such as the likelihood of a certain number of households owning at least one dog. By analyzing our random sample of 15 households out of millions in the U.S., we can make informed predictions and decisions regarding dog ownership trends. Using tools like the binomial distribution formula, we calculate specific probabilities, such as exactly 8 households having one dog or more than 10 households owning dogs. It transforms raw data into comprehensible insights.

Overall, statistical analysis is critical for understanding the data we collect and for validating hypotheses, thus making it a fundamental aspect of probability theory and real-world applications.