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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. 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 .

Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, as shown in the accompanying picture. A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red, and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments. a. When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? b. In the roulette wheel shown, black and red segments alternate. Suppose instead that all red segments were grouped together and that all black segments were together. Does this increase the probability that the ball will land in a red segment? Explain. c. Suppose that you watch 1000 spins of a roulette wheel and note the color that results from each spin. What would be an indication that the wheel was not balanced?

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E} .\) One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

An article in the New York Times reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=0.01\) for people who suffer cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and difficulty finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The basis for the New York Times article was a research study of 2,329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2,329 cardiac arrest sufferers do you think survived? Explain.
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