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Chapter 5: Problem 34

An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. a. Use the given probability information to set up a "hypothetical 1000 " table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. ii. the probability that a randomly selected customer purchases an extended warranty for neither the washer nor the dryer.

Short Answer

Expert verified
The probability that a randomly selected customer purchases an extended warranty for both the washer and the dryer is 40%. The probability that a randomly selected customer purchases an extended warranty for neither the washer nor the dryer is 41%.

Step by step solution

01

Setting up the hypothetical table

Let's assume there are 1000 customers. This gives us a total of 520 (52% of 1000) customers will purchase the extended warranty for the washer, 470 (47% of 1000) customers will purchase the extended warranty for the dryer, and 590 (59% of 1000) customers will purchase at least one of the two extended warranties. So the number of customers who buy the warranty for both appliances would be the sum of those who buy the washer and dryer warranty minus those who buy at least one, which gives us \(520 + 470 - 590 = 400\).
02

Find the probability for buying both warranties

To find the probability a customer buys both warranties, we divide the number of customers who buy both by the total number of customers. Therefore the probability equals \(400/1000 = 0.4\) or 40%.
03

Find the probability for buying neither warranty

To find the probability a customer buys no warranty, we can subtract those who bought at least one from the total customers. Therefore, the number of customers who buy neither warranty equals \(1000 - 590 = 410\). Then divide by the total number of customers, the probability thus equal to \(410/1000 = 0.41\) or 41%.

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Key Concepts

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

Extended Warranty
Extended warranties are additional protection plans that consumers can purchase to cover repairs or replacements beyond the standard manufacturer's warranty. These plans can often save money if appliances break down unexpectedly after the manufacturer's warranty expires. In our exercise, the focus is on whether customers who purchase washers and dryers also buy extended warranties for their appliances. The extended warranty upsell is an example of analyzing consumer behavior using probability to anticipate future sales trends.
Hypothetical Table
A hypothetical table is a statistical tool used to simplify complex probability calculations by assuming a sample size, often 1000 units, to represent customer behavior. This approach was utilized in the exercise to make calculations clearer and more manageable. Using a hypothetical 1000 customers, we calculated specific data points such as:
  • 520 customers purchasing the warranty for the washer
  • 470 customers purchasing the warranty for the dryer
  • 590 customers purchasing at least one warranty
By using these hypothetical numbers, we can find the overlap and distinctions in customer choices, leading to an easier understanding of the probability problems presented.
Statistical Problem Solving
Problem-solving in statistics often involves translating real-world scenarios into numerical and probabilistic contexts. In this exercise, statistical problem-solving allows us to determine the likelihood of different warranty purchase behaviors among customers. The steps included setting up a hypothetical table to organize our data and then performing calculations to find probabilities, like those buying both warranties or none at all. This method showcases the importance of structured data organization and logical reasoning in statistical analysis. Understanding the basics of creating hypothetical scenarios can significantly improve your problem-solving capabilities in statistics.

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Most popular questions from this chapter

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: \(P(\) Seed 1 defeats Seed 4\()=0.8\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 3)=0.7\) \(P(\) Seed 2 defeats \(\operatorname{Seed} 3)=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats Seed 4) \(=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

Lyme disease is the leading tick-borne disease in the United States and Europe. Diagnosis of the disease is difficult and is aided by a test that detects particular antibodies in the blood. The article "Laboratory Considerations in the Diagnosis and Management of Lyme Borreliosis" (American Journal of Clinical Pathology [1993]: \(168-174\) ) used the following notation: + represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities were reported in the article:\(\begin{aligned} P(L) &=0.00207 \\ P\left(L^{C}\right) &=0.99793 \\ P(+\mid L) &=0.937 \\ P(-\mid L) &=0.063 \\ P\left(+\mid L^{C}\right) &=0.03 \\ P\left(-\mid L^{C}\right) &=0.97 \end{aligned}\) a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a "hypothetical \(1000 "\) table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use the table constructed in Part (b) to calculate this probability.

Consider a chance experiment that consists of selecting a student at random from a high school with 3,000 students. a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.41\) and \(P(E)=0.23\). i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F) ?\) b. Suppose that for events \(A\) and \(B, P(A)=0.26, P(B)=0.34\), and \(P(A \cup B)=0.47 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

The probability of getting a king when a card is selected at random from a standard deck of 52 playing cards is \(\frac{1}{13}\). a. Give a relative frequency interpretation of this probability. b. Express the probability as a decimal rounded to three decimal places. Then complete the following statement: If a card is selected at random, I would expect to see a king about_____ times in 1000 .
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