/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A car dealer offers a \( 500\) d... [FREE SOLUTION] | 91Ó°ÊÓ

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A car dealer offers a \( 500\) discount to customers if they agree to buy a car immediately without doing further research. Suppose \(30%\) of all customers who visit him accept this offer. Depending on whether or not a given customer accepts the offer, let \(X\) be either 1 or 0 , respectively. a. If \(n=5\) customers, find the probability distribution of the proportion of customers who will accept the offer. (Hint: List all possible values for the sample proportion and their chances of occurring.) b. Referring to part a, what are the mean and standard deviation of the sample proportion? c. Repeat part b for a group of \(n=10\) customers and \(n=100\) customers. d. What happens to the mean and standard deviation of the sample proportion as \(n\) increases?

Short Answer

Expert verified
The mean remains 0.3 while the standard deviation decreases as \( n \) increases.

Step by step solution

01

Identify Binomial Distribution

This is a binomial probability problem where each customer is a trial, and they either accept the offer (success) or don't (failure). The probability of a customer accepting the offer is given as \( p = 0.3 \). The number of trials \( n \) is initially 5 customers.
02

Calculate Possible Outcomes

For part a, we calculate the probability distribution of the proportion of customers accepting the offer. For \( n = 5 \), the possible outcomes for the number of customers accepting the offer \( X \) are 0, 1, 2, 3, 4, and 5. The proportion \( \hat{p} = \frac{X}{n} \) will be 0, 0.2, 0.4, 0.6, 0.8, and 1 respectively.
03

Find Probability Mass Function

The probability mass function for each outcome is given by the binomial formula \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Calculate this for \( X = 0, 1, 2, 3, 4, \) and \( 5 \) to get the probabilities for each \( \hat{p} \).
04

Calculate Mean and Standard Deviation (n=5)

The mean of \( \hat{p} \) for \( n = 5 \) is \( E(\hat{p}) = p = 0.3 \). The standard deviation is \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.3 \times 0.7}{5}} \approx 0.2 \).
05

Calculate Mean and Standard Deviation (n=10)

The mean \( E(\hat{p}) = p = 0.3 \) remains the same for \( n = 10 \). The standard deviation is \( \sigma_{\hat{p}} = \sqrt{\frac{0.3 \times 0.7}{10}} \approx 0.148 \).
06

Calculate Mean and Standard Deviation (n=100)

For \( n = 100 \), the mean \( E(\hat{p}) = p = 0.3 \). The standard deviation is \( \sigma_{\hat{p}} = \sqrt{\frac{0.3 \times 0.7}{100}} \approx 0.046 \).
07

Impact of Increasing n on Mean and Standard Deviation

As \( n \) increases, the mean \( E(\hat{p}) \) remains constant at 0.3. However, the standard deviation \( \sigma_{\hat{p}} \) decreases, indicating that the sample proportion becomes more accurate and concentrated around the mean as the sample size increases.

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Key Concepts

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

Probability Distribution
In probability theory and statistics, a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes. In the context of our exercise, we are dealing with a binomial distribution, which is a type of probability distribution. This occurs in situations where there are fixed probabilities of success and failure, like the car dealership example where customers either accept a discount offer (success) or not (failure).

Here, each customer represents a trial, and the probability (\(p\)) that a customer will accept the discount is 0.3. A probability distribution helps us determine the likelihood of observing various numbers of customers accepting the discount. For instance, if 5 customers are considered, each has a chance of 30% to accept the offer, leading to a distribution where possible success outcomes can range from 0 to 5 accepting customers.
  • A probability mass function can be used in discrete distributions to calculate the probability of each possible outcome.
  • This distribution helps us anticipate and understand the variability in observed processes from even small groups of trials.
Sample Proportion
The sample proportion, denoted as \(\hat{p}\), is the fraction of total observations in our sample that have a particular characteristic. In our exercise involving customers, this proportion represents the number of customers accepting the offer out of the total surveyed. It is calculated using \(\hat{p} = \frac{X}{n}\), where \(X\) is the number of successes, and \(n\) is the total number of trials.

For a group of 5 customers, possible \(\hat{p}\) values could be \[0, 0.2, 0.4, 0.6, 0.8, 1\], reflecting 0 to 5 customers accepting the offer respectively. The sample proportion is crucial because it measures success in experiments and surveys, providing a snapshot of behavior within the sampled group.
  • Sample proportion gives a practical understanding of how closely our sample reflects the real-world probability.
  • It helps predict future outcomes by detailing current sample behaviors.
Mean and Standard Deviation
The mean is a measure of central tendency, showing the average outcome we can expect in our sample. For a binomial distribution, the mean of the sample proportion \(E(\hat{p})\) remains constant at the population probability \(p\). In this example, regardless of whether we survey 5, 10, or 100 customers, the mean sample proportion remains 0.3, matching the expected probability of a customer accepting the offer.

The standard deviation offers insights into the variability or spread of the sample outcomes around this mean. It is calculated using \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Essentially, it tells us how much the sample proportions fluctuate around the mean. With \(n=5\), the standard deviation is approximately 0.2. As \(n\) increases to 10 or 100, this standard deviation becomes 0.148 and 0.046 respectively.
  • This decrease in standard deviation with increased sample size indicates less variation, making the sample proportion more precise.
  • Understanding both these measures helps estimate the accuracy and reliability of our sample in representing the broader population.
Impact of Sample Size
Sample size, denoted as \(n\), plays a crucial role in statistical analysis. In the context of our problem, increasing the number of customers surveyed (\(n\)) has significant effects on the observed sample proportion attributes. Although the mean \(E(\hat{p})\) remains constant regardless of \(n\), the precision and reliability of our estimation improve.

As \(n\) increases, the standard deviation \(\sigma_{\hat{p}}\) declines, showing that larger samples provide more stable estimates with less variation. This phenomenon highlights why larger samples are preferable— they yield results that more closely reflect the actual population conditions, with sample proportions clustering tightly around the mean.
  • Increased sample size reduces sampling error, ensuring that the sample outcome better represents the true population parameter.
  • Larger samples, therefore, provide more confident and precise predictions about future or general behavior in similar scenarios.

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Most popular questions from this chapter

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