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Despite reports that dark chocolate is beneficial to the heart, \(47 \%\) of adults still prefer milk chocolate to dark chocolate. \({ }^{11}\) Suppose a random sample of \(n=5\) adults is selected and asked whether they prefer milk chocolate to dark chocolate. a. What is the probability that all five adults say that they prefer milk chocolate to dark chocolate? b. What is the probability that exactly three of the five adults say they prefer milk chocolate to dark chocolate? c. What is the probability that at least one adult prefers milk chocolate to dark chocolate?

Short Answer

Expert verified
Answer: The probability that all five adults prefer milk chocolate to dark chocolate is approximately \(0.0401\), that exactly three of the five adults prefer milk chocolate to dark chocolate is approximately \(0.3623\), and that at least one adult prefers milk chocolate to dark chocolate is approximately \(0.9579\).

Step by step solution

01

a. Probability that all five adults prefer milk chocolate to dark chocolate

We need to find the probability that all 5 adults in the sample prefer milk chocolate over dark chocolate (\(k = 5\)). We will use the binomial probability formula: \(P(5) = \begin{pmatrix} 5\\ 5\end{pmatrix} \cdot (0.47)^5 (1-0.47)^{5-5}\) Calculate the probability: \(P(5) = 1 \cdot (0.47)^5 (0.53)^0 = (0.47)^5\) \(P(5) \approx 0.0401\) The probability that all five adults prefer milk chocolate to dark chocolate is approximately \(0.0401\).
02

b. Probability that exactly three of the five adults prefer milk chocolate to dark chocolate

We need to find the probability that exactly 3 adults in the sample prefer milk chocolate over dark chocolate (\(k = 3\)). We will use the binomial probability formula: \(P(3) = \begin{pmatrix} 5\\ 3\end{pmatrix} \cdot (0.47)^3 (1-0.47)^{5-3}\) Calculate the probability: \(P(3) = 10 \cdot (0.47)^3 (0.53)^2 \approx 0.3623\) The probability that exactly three of the five adults prefer milk chocolate to dark chocolate is approximately \(0.3623\).
03

c. Probability that at least one adult prefers milk chocolate to dark chocolate

We need to find the probability that at least one adult prefers milk chocolate (\(k \geq 1\)). To do this, we can find the complementary probability that none of the adults prefer milk chocolate (\(k = 0\)) and subtract it from 1. We will use the binomial probability formula: \(P(0) = \begin{pmatrix} 5\\ 0\end{pmatrix} \cdot (0.47)^0 (1-0.47)^{5-0}\) Calculate the probability: \(P(0) = 1 \cdot 1 \cdot (0.53)^5 \approx 0.0421\) Now, find the complementary probability: \(P(K \geq 1) = 1 - P(0) = 1 - 0.0421\approx 0.9579\) The probability that at least one adult prefers milk chocolate to dark chocolate is approximately \(0.9579\).

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Key Concepts

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

Probability
Probability is a tool that helps us quantify uncertainty. It gives us a numerical value between 0 and 1, representing the likelihood of an event happening. For example, if an event has a probability of 1, it is certain to happen, while a probability of 0 indicates that the event is impossible. In the case of the chocolate preference example with adults, we use probabilities to determine the chances of different scenarios occurring during our survey.

In the exercise, we explore the probability of adults preferring milk chocolate over dark chocolate. We express these scenarios using a value derived from the ratio of individuals who prefer milk chocolate over the total number of surveyed adults. Calculating such a possibility helps in making informed decisions and predictions based on varying outcomes.
Random Sample
A random sample is a key concept in statistics. It ensures that every individual in a population has an equal chance of being selected for the study. In the context of the chocolate preference survey, obtaining a random sample of five adults helps depict an accurate representation of the whole population.

Random sampling aims to minimize biases and allow for manageable and reliable predictions. The random nature of sampling means that inherent personal preferences don't skew results, maintaining the study’s integrity. It helps avoid patterns that could misrepresent the broader population's chocolate preferences.
Complementary Probability
Complementary probability is a handy concept in probability, used to calculate the chance of an event not occurring. Simply put, if you have the probability of an event happening, the complementary probability helps you deduce the probability of the event not happening by subtracting it from 1.

In our exercise, we calculated the probability that zero adults prefer milk chocolate. To find the probability of at least one adult preferring milk chocolate, we used complementary probability. We subtracted the probability of none preferring it (i.e., all preferring dark chocolate) from 1, giving us the probability where at least one adult would opt for milk chocolate. This approach streamlines calculations involving the occurrence of at least one particular outcome, offering an efficient method to find such probabilities.
Binomial Probability Formula
The binomial probability formula serves as a vital tool in probability calculations, especially when dealing with binary outcomes like yes/no, success/failure, or in this case, preference for one type of chocolate over another. This formula is defined as: \[P(k) = \begin{pmatrix} n\k\end{pmatrix} \cdot p^k \cdot (1-p)^{n-k} \] where
  • \(n\) is the sample size,
  • \(k\) is the number of successful outcomes we are interested in,
  • \(p\) is the probability of a single success (adult preferring milk chocolate),
  • and \((1-p)\) is the probability of failure.

We used this formula when calculating the probabilities for different numbers of adults preferring milk chocolate in our sample. It's a versatile tool, perfect for scenarios where there are two potential outcomes, allowing researchers to calculate specific probabilities with precision.

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Most popular questions from this chapter

A packaging experiment is conducted by placing two different package designs for a breakfast food side by side on a supermarket shelf. On a given day, 25 customers purchased a package of the breakfast food from the supermarket. Let \(x\) equal the number of buyers who choose the second package design. a. If there is no preference for either of the two designs, what is the value of \(p,\) the probability that a buyer chooses the second package design? b. If there is no preference, use the results of part a to calculate the mean and standard deviation of \(x\) c. If 5 of the 25 customers choose the first package design and 20 choose the second design, what do you conclude about the customers' preference for the second package design?

The recession has caused many people to use their credit cards far less. In fact, in the United States, \(60 \%\) of consumers say they are committed to living with fewer credit cards. \({ }^{15} \mathrm{~A}\) sample of \(n=400\) consumers with credit cards are randomly selected. a. What is the average number of consumers in the sample who said they are committed to living with fewer credit cards? b. What is the standard deviation of the number in the sample who said they are committed to living with fewer credit cards? c. Within what range would you expect to find the number in the sample who said they are committed to living with fewer credit cards? d. If only 200 of the sample of consumers said they were committed to living with fewer credit cards, would you consider this unusual? Explain. What conclusion might you draw from this sample information?

A market research firm hires operators to conduct telephone surveys. The computer randomly dials a telephone number, and the operator asks the respondent whether or not he has time to answer some questions. Let \(x\) be the number of telephone calls made until the first respondent is willing to answer the operator's questions. Is this a binomial experiment? Explain.

A West Coast university has found that about \(90 \%\) of its accepted applicants for enrollment in the freshman class will actually enroll. In 2012,1360 applicants were accepted to the university. Within what limits would you expect to find the size of the freshman class at this university in the fall of \(2012 ?\)

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

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