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Chapter 3: Problem 75

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 debit: 20, Variance: 16; Mean for not cash: 70, Variance: 21.

Step by step solution

01

Understanding the Problem

Firstly, we want to find the mean and variance of the number of customers who pay with a debit card among 100 customers. We are given that the probability a customer pays with a debit card (B) is \(P(B) = 0.2\).
02

Identifying the Distribution

The scenario described follows a binomial distribution, where we have a fixed number of trials (100 customers), each with two possible outcomes (paying with debit card or not). The probability of paying with a debit card is 0.2.
03

Calculating the Mean (a)

For a binomial distribution, the mean \(\mu\) is given by \(\mu = n \cdot p\), where \(n = 100\) and \(p = 0.2\). Thus, the mean number of customers paying with a debit card is \(100 \cdot 0.2 = 20\).
04

Calculating the Variance (a)

The variance \(\sigma^2\) for a binomial distribution is computed as \(\sigma^2 = n \cdot p \cdot (1-p)\). Substituting the known values: \(\sigma^2 = 100 \cdot 0.2 \cdot 0.8 = 16\).
05

Addressing Part (b)

Now, for part (b), we need the mean and variance for customers who don't pay with cash. They can pay with either a credit card or a debit card. Thus, \(P(\text{not cash}) = P(A) + P(B) = 0.5 + 0.2 = 0.7\).
06

Calculating the Mean (b)

Using the binomial distribution for not paying with cash, mean \(\mu = n \cdot p = 100 \cdot 0.7 = 70\).
07

Calculating the Variance (b)

The variance \(\sigma^2 = n \cdot p \cdot (1-p) = 100 \cdot 0.7 \cdot 0.3 = 21\).

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Key Concepts

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

Probability
In the context of this problem, probability refers to the likelihood that a particular event, such as a customer paying with a debit card, occurs. There are three payment methods: credit card, debit card, and cash, each with a specific probability.
Let's break down the probabilities:
  • Probability of paying with a credit card, \(P(A)\), is 0.5.
  • Probability of paying with a debit card, \(P(B)\), is 0.2.
  • Probability of paying with cash, \(P(C)\), is 0.3.
These probabilities must sum up to 1, confirming they cover all possibilities. When we assess the likelihood of events, like how many customers use a debit card, we deal with binomial probability because there's a defined set of trials (100 customers) and each trial's outcome is either a success (debit payment) or a failure (other payment). Understanding this helps in calculating associated means and variances efficiently.
Mean and Variance
The concepts of mean and variance are essential for understanding the distribution of outcomes in a binomial setting. They help quantify the expected results from repeated trials and the variation around that expectation.

Mean

In the binomial distribution, the mean \(\mu\) represents the average number of successful outcomes, such as payments made via debit card, over many trials. It's calculated as \(\mu = n \cdot p\), where \(n\) is the number of trials (customers) and \(p\) is the probability of success per trial (debit payment). For 100 customers with a debit payment probability of 0.2, the mean is \(100 \cdot 0.2 = 20\). This means, on average, 20 out of 100 customers will pay with a debit card.

Variance

Variance \(\sigma^2\) measures the extent to which the number of debit card payments varies from the mean. In a binomial distribution, variance is calculated using \(\sigma^2 = n \cdot p \cdot (1-p)\). For the debit card example, this is \(100 \cdot 0.2 \cdot 0.8 = 16\). This indicates how spread out the numbers of debit card payments can be around the mean.
Independent Trials
In probability, independent trials imply that the outcome of one trial does not influence the outcome of another. When we say customers make independent choices for payment, each customer's choice is unaffected by others' choices.

Importance in Binomial Distribution

For a binomial distribution, assuming independent trials is crucial. It ensures that the probability of success (like choosing to pay with a debit card) remains constant throughout each trial (or customer interaction). This characteristic allows us to apply formulas for mean and variance accurately.
The independence simplifies calculations and helps ascertain predictions without bias from previous trials. Consequently, even with distinct trials (e.g., customers with different payment preferences), we can rely on calculated probabilities and outcomes over time.

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Most popular questions from this chapter

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{c|cccccc} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]

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Suppose that only \(10 \%\) of all computers of a certain type experience CPU failure during the warranty period. Consider a sample of 10,000 computers. a. What are the expected value and standard deviation of the number of computers in the sample that have the defect? b. What is the (approximate) probability that more than 10 sampled computers have the defect? c. What is the (approximate) probability that no sampled computers have the defect?

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

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