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According to the Humane Society of America, there are approximately 77.8 million owned dogs in the United States, and approximately \(50 \%\) of dog- owning households have small dogs. \({ }^{8}\) Suppose the \(50 \%\) figure is correct and that \(n=15\) dog-owning households are randomly selected for a pet ownership survey. a. What is the probability that exactly eight of the households have small dogs? b. What is the probability that at most four of the households have small dogs? c. What is the probability that more than 10 households have small dogs?

Short Answer

Expert verified
a. Exactly 8 of the 15 selected households have small dogs. b. At most 4 of the 15 selected households have small dogs. c. More than 10 of the 15 selected households have small dogs.

Step by step solution

01

Define the parameters

The parameters in this problem are: - number of trials (households), n = 15 - probability of having a small dog, p = 0.5
02

Calculate the probability of exactly 8 households having small dogs

To calculate this probability, we will use the binomial probability formula with k = 8: \[ P(k=8) = C(15,8)\times 0.5^{8} \times (1-0.5)^{15-8} \] Compute the binomial coefficient and the probabilities in the expression: \[ P(k=8) = 6435 \times 0.5^{8} \times 0.5^{7} \] Now calculate the final probability: \[ P(k=8) \approx 0.1964 \]
03

Calculate the probability of at most 4 households having small dogs

To calculate this probability, we need to find the sum of the probabilities P(k) for k=0, k=1, k=2, k=3, and k=4: \[ P(k \le 4) = \sum_{k=0}^{4} C(15,k)\times 0.5^{k} \times (1-0.5)^{15-k} \] Compute the binomial coefficients and then compute each probability in the sum: \[ P(k \le 4) \approx 0.0132 \]
04

Calculate the probability that more than 10 households have small dogs

To calculate this probability, we need to find the sum of the probabilities P(k) for k=11, k=12, k=13, k=14, and k=15: \[ P(k > 10) = \sum_{k=11}^{15} C(15,k)\times 0.5^{k} \times (1-0.5)^{15-k} \] Compute the binomial coefficients and then compute each probability in the sum: \[ P(k > 10) \approx 0.0212 \] Thus, the probabilities are: a. Probability that exactly 8 households have small dogs: 0.1964 b. Probability that at most 4 households have small dogs: 0.0132 c. Probability that more than 10 households have small dogs: 0.0212

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Key Concepts

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

Binomial Probability Distribution
When addressing problems involving a fixed number of trials with two possible outcomes—often termed 'success' and 'failure'—the binomial probability distribution serves as an essential tool. It's the go-to method for determining the likelihood of a given number of successes over several trials, assuming each trial is independent and the probability of success stays constant.

In our exercise, owning a small dog is considered a 'success' each household is an independent trial. With \( n = 15 \) households and a success probability (\( p = 0.5 \)), we're situated perfectly to use the binomial distribution for our calculations.

The binomial probability of exactly \( k \) successes is given by:
  • \( P(k) = C(n,k) \times p^{k} \times (1-p)^{n-k} \) where \( C(n, k) \) signifies the number of combinations or ways to choose \( k \) successes from \( n \) trials.
To find out the probability of exactly eight households having small dogs, we employ the formula and substitute the relevant values. This approach reflects the essence of the binomial distribution, portraying how it simplifies analyzing scenarios with binary outcomes distributed across multiple events.
Probability Calculations
Probability calculations in scenarios like these can sometimes seem daunting, but by using the proper approach, they become quite manageable. The key is understanding how to apply formulas and interpret the outcomes—a critical skill in the realm of probability and statistics.

For instance, the probability of exactly one event occurring is straightforward—it's simply \( P(k) \) with the corresponding number of successes. However, calculating probabilities across a range of values, such as 'at most four households,' requires summing individual probabilities for \( k = 0 \) to \( k = 4 \). We demonstrated this in the exercise, highlighting a practical application of summing probabilities to capture a broader outcome.

Moreover, when the question concerns the likelihood of 'more than 10 households,' we turn our attention to the upper end of the distribution. Here we sum the probabilities from \( k = 11 \) to \( k = 15 \), showcasing two sides of the same probabilistic coin: sums of discrete probabilities over a range. Such calculations are fundamental in making predictions and informed decisions based on probabilities.
Random Variables
A random variable is essentially a function that assigns numerical values to the outcomes of random phenomena. In the context of our exercise, the number of households with small dogs is a discrete random variable—where 'discrete' refers to variables that take on a countable number of distinct values.

In binomial experiments, the random variable of interest typically counts the number of successes, which in our scenario corresponds to the number of households owning small dogs. This variable can assume any whole number from zero to the total number of trials, which aligns with the essence of a binomial random variable.

Understanding random variables is crucial for interpreting real-world situations in mathematical terms, facilitating the application of probability distributions, and ultimately, quantifying uncertainty. It is the cornerstone of probability theory and is extensively employed in statistical analysis and hypothesis testing.

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Most popular questions from this chapter

Let \(x\) be a hypergeometric random variable with \(N=15, n=3,\) and \(M=4\). Use this information to answer the questions in Exercises 14-17. What portion of the population of measurements fall into the interval \((\mu \pm 2 \sigma) ?\) Into the interval \((\mu \pm 3 \sigma) ?\) Do the results agree with Tchebysheff's Theorem?

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