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Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 payments were made, and 37,100 of them were by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

Short Answer

Expert verified
The estimated probability that a randomly selected customer will pay by credit card is approximately 0.46 or 46%.

Step by step solution

01

Identify the total number of outcomes

The total number of outcomes is the total payments made at the grocery store which is 80,500.
02

Identify the number of favorable outcomes

The number of favorable outcomes is the number of payments made by credit card which is 37,100.
03

Calculate the Probability

The probability of a randomly selected customer paying by credit card is calculated by dividing the number of favorable outcomes by the total number of outcomes. Using these numbers, the calculation would be \(\frac{37100}{80500}\).

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Key Concepts

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

Random Sampling
Random sampling is a fundamental method used in the field of statistics to select a subset of individuals or observations from within a statistical population to estimate characteristics of the whole population. Understanding this concept is critical when it comes to tackling real-world problems where it's impractical or impossible to examine an entire population.

In the exercise provided, the grocery store's payments represent the population, and each payment is an individual observation. By collecting data from the entire three months, which involves 80,500 payments, the store ensures that the sample reflects the broad range of customers. The concept assumes that every payment had an equal chance of being included in the data set, which is a key principle of random sampling. It minimizes biases that can skew results and provides a reliable foundation for estimating the probability of future events, such as the likelihood of customers paying with credit cards.

Random sampling's real value lies in its ability to provide accurate estimations that can be generalized to the larger group, which is especially useful when the total population is too large to analyze fully.
Favorable Outcomes
Whether you are calculating the likelihood of rolling a six on a die or estimating the probability of an event in real life, the concept of favorable outcomes is central to probability. Favorable outcomes are those specific results that we're interested in when performing a probability experiment.

In the context of our grocery store example, a favorable outcome refers to an event where a customer pays by credit card. Out of the 80,500 total payments, 37,100 were made with a credit card. These 37,100 payments are our favorable outcomes. They are 'favorable' simply because these are the outcomes we are counting when we aim to estimate the probability of a particular occurrence.

Understanding what constitutes a favorable outcome is essential for accurate probability calculation. It affects the numerator in our probability fraction and thus directly impacts the calculated likelihood of an event occurring. Additionally, providing clarity on what exactly are the favorable outcomes can help students easily grasp the concept, specially in the context of complex, real-world problems.
Probability Calculation
The probability calculation is the mathematical process used to find the likelihood of a particular event happening. This is expressed as a number between 0 and 1, where 0 indicates an impossibility, and 1 represents certainty.

Following the steps solved in the exercise, calculating the probability involves dividing the number of favorable outcomes by the total number of possible outcomes. As seen in the grocery store scenario, the probability that a customer pays by credit card is the quotient of the favorable credit card payments (37,100) over the total payments made (80,500). Therefore, using the formula for probability, \( P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} \) where A is the event of a customer paying by credit card, we get \( P(A) = \frac{37100}{80500} \) which simplifies to a decimal that estimates the probability of the event.

The calculation can be interpreted as the expected frequency of the event occurring in a long series of trials. Probability calculations are vital for decision making in many fields, enabling businesses and individuals to anticipate likely outcomes and plan accordingly.

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Most popular questions from this chapter

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.

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.

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.

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?

If you were to roll a fair die 1,000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

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