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Chapter 3: Problem 65

Customers at a gas station pay with a credit card (A), debit card \((B)\), or cash \((C)\). Assume that successive customers make independent choices, with \(P(A)=.5\), \(P(B)=.2\), and \(P(C)=.3\). a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

Short Answer

Expert verified
Mean for part a is 20; variance is 16. Mean for part b is 70; variance is 21.

Step by step solution

01

Identify the Distribution for Part a

This problem involves finding the number of customers who pay with a debit card. We can model this situation using a binomial distribution because each customer either pays with a debit card or doesn't, making the outcome binary. Let \( X \) be the number of customers paying with a debit card, so \( X \sim \text{Binomial}(n=100, p=0.2) \).
02

Calculate the Mean for Part a

In a binomial distribution, the mean \( \mu \) is given by \( np \). Here, \( n = 100 \) and \( p = 0.2 \). Calculate the mean as \( \mu = 100 \times 0.2 = 20 \).
03

Calculate the Variance for Part a

For a binomial distribution, the variance \( \sigma^2 \) is given by \( np(1-p) \). Substitute \( n = 100 \), \( p = 0.2 \), and \( 1-p = 0.8 \) to find the variance as \( \sigma^2 = 100 \times 0.2 \times 0.8 = 16 \).
04

Reformulate the Question for Part b

In part (b), we need the statistics for the number of customers who don't pay with cash. This is equivalent to finding those who pay with either a credit or debit card, combining these two categories. Let \( Y \) represent the number of non-cash paying customers, where \( Y \sim \text{Binomial}(n=100, p') \) and \( p' = P(A) + P(B) = 0.5 + 0.2 = 0.7 \).
05

Calculate the Mean for Part b

Using the binomial distribution mean formula, \( \mu' = np' \). Substitute \( n = 100 \) and \( p' = 0.7 \) to calculate the mean \( \mu' = 100 \times 0.7 = 70 \).
06

Calculate the Variance for Part b

The variance for this new binomial distribution is \( \sigma'^2 = np'(1-p') \). Substitute \( n = 100 \), \( p' = 0.7 \), and \( 1 - p' = 0.3 \) to obtain the variance \( \sigma'^2 = 100 \times 0.7 \times 0.3 = 21 \).

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Key Concepts

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

Mean of Binomial Distribution
When dealing with a binomial distribution, understanding the mean is crucial. The mean of a binomial distribution helps us understand the expected number of successes in a given number of trials. In simpler terms, it's like predicting how many heads you will get if you flip a coin several times.

The mean is calculated using the formula \( \mu = np \), where \( n \) is the total number of trials, and \( p \) is the probability of success on each trial. For example, if you have 100 customers at a gas station, and each customer has a 0.2 probability of paying with a debit card, then the mean number of customers paying with a debit card can be calculated as \( 100 \times 0.2 = 20 \). This means, on average, you can expect 20 customers to pay with a debit card among those 100 customers.

This concept is essential in predicting outcomes and making plans based on expectations. By knowing the mean, businesses can predict trends and prepare accordingly.
Variance of Binomial Distribution
While the mean tells us the expected number of successes, the variance gives us an idea of how much fluctuation or variation there is from that expectation. In a binomial distribution, the variance measures how spread out the numbers are.

The formula for calculating the variance of a binomial distribution is \( \sigma^2 = np(1-p) \). Here, \( n \) is the number of trials, \( p \) is the probability of success, and \( 1-p \) represents the probability of failure. In our gas station example, using the debit card payment with \( p = 0.2 \) and \( n = 100 \), the variance can be calculated as \( 100 \times 0.2 \times 0.8 = 16 \).

A variance of 16 implies that the number of customers using a debit card will usually be close to the mean of 20, but can vary based on real-world conditions. Understanding variance helps businesses assess risk and ensure they are prepared for fluctuations in customer behavior.
Independent Choices in Statistics
In statistics, making assumptions about independence is often key to solving problems. Independent choices mean that the outcome of one event doesn't affect the outcome of another. For the gas station scenario, assuming that each customer's choice of payment method does not influence the next customer's choice simplifies our calculations.

This independence allows us to utilize the binomial distribution. Under this assumption, each payment choice is a separate trial with a known probability. For example, whether one customer pays by debit does not change the chances of the next customer paying by credit or cash.

Recognizing and applying the concept of independent choices ensures that statistical models accurately reflect real-world situations. It contributes to forming predictive analyses, which help in decision-making and strategizing in various fields, from retail operations to scientific research.

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Most popular questions from this chapter

A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is \begin{tabular}{l|cccc} \(y\) & 0 & 1 & 2 & 3 \\ \hline\(p(y)\) & \(.60\) & \(.25\) & \(.10\) & \(.05\) \end{tabular} a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students \((1,2,3\), and 4) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated as \((2,4,3,1)\). a. List the other 23 possible outcomes. b. Let \(X\) denote the number of students who receive their own book. Determine the pmf of \(X\).

Write a general rule for \(E(X-c)\) where \(c\) is a constant. What happens when \(c=\mu\), the expected value of \(X\) ?
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