/*! 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 16 Suppose you want to estimate the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 16

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 purchases were made, and 37,100 of them were paid for 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 46.09%.

Step by step solution

01

Note down the given numbers

We are given that there were 37,100 credit card purchases and a total of 80,500 purchases. So, let's denote the number of credit card purchases as CC and the total number of purchases as TP. CC = 37,100 TP = 80,500
02

Calculate the proportion of credit card purchases

To find the estimated probability that a randomly selected customer will pay by credit card, we need to divide the number of credit card purchases by the total number of purchases: Probability = \( \frac{CC}{TP} \)
03

Plug in the given numbers and calculate the probability

Now we simply substitute the given numbers into the formula: Probability = \( \frac{37,100}{80,500} \) By calculating this, we find: Probability = 0.4609 (rounded to four decimal places)
04

Express the probability as a percentage

To express the probability as a percentage, we simply multiply the decimal probability by 100: Percentage = Probability 脳 100 Percentage = 0.4609 脳 100 = 46.09% So, the estimated probability that a randomly selected customer will pay by credit card is approximately 46.09%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Credit Card Payments
Credit card payments represent a significant portion of transactions in many retail environments. In our exercise, we focus on the estimation of the likelihood that a customer will choose to pay using a credit card at a grocery store. This form of payment is popular due to its convenience and security. Credit card payments allow customers to purchase goods without the immediate need for cash.

It's important to consider factors that can affect a customer's choice of payment method, such as the store's location, the demographics of its customers, promotional offers, and more. Understanding these factors can provide businesses with insights into consumer behavior and trends, helping them plan more effectively and provide better service.
Estimate Probability
Estimating probability is an essential concept in statistics that helps us predict the likelihood of an event occurring. In the context of our exercise, we're interested in the probability that a customer at a grocery store will pay with a credit card. This helps in making decisions related to business strategy and customer management.

To estimate probability, we use a simple mathematical formula. We divide the number of successful events (in this case, payments made by credit card) by the total number of events (total purchases). The formula is:
\[ Probability = \frac{CC}{TP} \]where CC is the number of credit card payments and TP is the total number of payments.

Calculating this simple ratio gives us an insight into customer behavior patterns regarding payment methods. Understanding probabilities can be pivotal for businesses in forecasting and resource allocation.
Proportion Calculation
Proportion calculation is a straightforward yet powerful tool used to determine the relative frequency of a particular occurrence within a larger context. In our exercise, we calculate the proportion of credit card payments out of total purchases.

A proportion can be calculated using the formula:Proportion = \( \frac{Part}{Whole} \)
In this context, the part is the number of credit card transactions (37,100), and the whole is the total number of purchases (80,500).

By dividing these numbers, we find the proportion: \( \frac{37,100}{80,500} = 0.4609 \). This means that approximately 46.09% of all transactions were completed using a credit card.

Understanding proportions is crucial as it allows businesses to interpret data in a way that is intuitive and easy to communicate, making it easier to identify trends and patterns in consumer behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to The Chronicle for Higher Education Almanac (2016), there were 1,003,329 Associate degrees awarded by U.S. community colleges in the \(2013-2014\) academic year. A total of 613,034 of these degrees were awarded to women. a. If a person who received an Associate degree in \(2013-\) 2014 is selected at random, what is the probability that the selected person will be female? b. What is the probability that the selected person will be male?

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 student council for a school of science and math has one representative from each of five academic departments: Biology (B), Chemistry (C), Mathematics (M), Physics (P), and Statistics (S). Two of these students are to be randomly selected for inclusion on a university-wide student committee. a. What are the 10 possible outcomes? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each outcome? c. What is the probability that one of the committee members is the statistics department representative? d. What is the probability that both committee members come from laboratory science departments?

5.62 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. In Exercise \(5.34,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. 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. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer.

A study of the impact of seeking a second opinion about a medical condition is described in the paper "Evaluation of Outcomes from a National Patient- Initiated Second-Opinion Program". Based on a review of 6791 patient-initiated second opinions, the paper states the following: "Second opinions often resulted in changes in diagnosis (14.8\%), treatment \((37.4 \%),\) or changes in both \((10.6 \%)\)." Consider the following two events: \(D=\) event that second opinion results in a change in diagnosis \(T=\) event that second opinion results in a change in treatment a. What are the values of \(P(D), P(T),\) and \(P(D \cap T) ?\) b. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(D\) and \(D^{C}\) and rows corresponding to \(T\) and \(T^{C}\). c. What is the probability that a second opinion results in neither a change in diagnosis nor a change in treatment? d. What is the probability that a second opinion results is a change in diagnosis or a change in treatment?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.