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According to Benford's law (Exercise \(5,\) page 359\(),\) the probability that the first digit of the amount of a randomly chosen invoice is an 8 or a 9 is \(0.097 .\) Suppose you examine randomly selected invoices from a vendor until you find one whose amount begins with an 8 or a \(9 .\) (a) How many invoices do you expect to examine until you get one that begins with an 8 or 9 ? Justify your answer. (b) In fact, you don't get an amount starting with an 8 or 9 until the 40 th invoice. Do you suspect that the invoice amounts are not genuine? Compute an appropriate probability to support your answer.

Short Answer

Expert verified
(a) 10.31 invoices; (b) Yes, with a probability of 0.00282, the 40th occurrence is unlikely.

Step by step solution

01

Understand the Problem

We need to solve two parts: find the expected number of invoices to sample before finding one that begins with an 8 or 9, and determine if finding such an invoice at the 40th attempt is unusual. We will use the concept of a geometric distribution for both parts.
02

Calculate Expected Number of Invoices

The problem is modeled by a geometric distribution where the probability of success (the invoice starts with an 8 or 9) is 0.097. For a geometric distribution, the expected number of trials until the first success is given by \( \frac{1}{p} \). Thus, the expected number of invoices is \( \frac{1}{0.097} \approx 10.31 \) invoices.
03

Identify Observed Outcome

You found the first invoice with 8 or 9 as the initial digit at the 40th invoice, significantly higher than the expected 10.31 invoices according to the geometric distribution based on Benford's Law.
04

Calculate Probability of Observed Outcome

The probability of waiting 39 invoices without success and then finding success on the 40th is given by the probability mass function of the geometric distribution: \( P(X = 40) = (1 - p)^{39} \cdot p \). Substituting \( p = 0.097 \), we get \((0.903)^{39} \cdot 0.097 \approx 0.00282\).
05

Interpret the Probability

The probability of observing the first invoice starting with an 8 or a 9 on the 40th attempt is approximately 0.00282, which is very low. This suggests that such an outcome is unusual under Benford's Law, which might imply that invoice amounts are not following the expected distribution.

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Key Concepts

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

Understanding the Geometric Distribution
The geometric distribution is a key concept when dealing with sequences of independent trials. It applies when you're trying to find the number of attempts needed to get the first success in a series of Bernoulli trials. In this context, each time you check an invoice, it is one trial, and a success occurs when the invoice starts with an 8 or 9.
A geometric distribution assumes that each trial is independent and the probability of success on each trial remains constant. The formula for the expected number of trials until the first success is simple: \( \frac{1}{p} \), where \( p \) is the probability of success.
  • For example, if the probability \( p = 0.097 \), you would expect about \( \frac{1}{0.097} \approx 10.31 \) trials before the first success.
This makes the geometric distribution particularly useful in situations where you are interested in waiting times or the number of attempts to achieve a success.
Probability as a Measure of Likelihood
Probability is a measure of how likely an event is to occur, ranging from 0 (impossible) to 1 (certain). In this exercise, you're dealing with the probability that an invoice starts with the digit 8 or 9.
Understanding probabilities helps us evaluate outcomes and make predictions. In this scenario, the probability of success on each invoice check is given as 0.097, based on Benford's Law.
  • Probabilities are foundational in calculating expectations with geometric distribution.
  • The context of the task involves assessing rare and common events.
This knowledge allows us to further explore if the observed number of trials (40 invoices) aligns with typical expectations or if there’s a reason to suspect an anomaly.
Calculating Expected Value in Geometric Distribution
In statistics, the expected value gives a measure of the center of a probability distribution. When we talk about the expected value in a geometric distribution, it indicates the average number of attempts you'll need before achieving success.
  • For a geometric distribution with success probability \( p \), the expected value is \( \frac{1}{p} \).
  • For this problem, with \( p = 0.097 \), the expected value is \( 10.31 \).
This insight is crucial because it provides a benchmark for understanding whether observed outcomes, like needing 40 attempts instead, are typical or outliers. The comparison provides groundwork for investigating statistical anomalies.
Spotting Statistical Anomalies
Statistical anomalies are observations that deviate significantly from what is expected based on given data or predicted by statistical models. Identifying anomalies is key in discovering underlying issues or errors in data.
  • In our problem, the finding of a "success" at 40 invoices, when you expect one roughly after 10.31 attempts, raises concerns.
  • The probability of such a result (success on the 40th attempt) is only about 0.00282, making it a rare event.
When working with data that should adhere to principles like Benford's Law, such low probabilities prompt us to investigate further. It might indicate errors in data, systematic biases, or fraudulent activities.

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Most popular questions from this chapter

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns, like too many round numbers, are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{7}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ (a) Show that this is a legitimate probability distribution. (b) Make a histogram of the probability distribution. Describe what you see. (c) Describe the event \(X \geq 6\) in words. What is \(P(X \geq 6) ?\) (d) Express the event "first digit is at most 5 " in terms of X. What is the probability of this event?

"Large Counts condition To use a Normal distribution to approximate binomial probabilities, why do we require that both \(n p\) and \(n(1-p)\) be at least \(10 ?\)

Exercises 27 to 29 refer to the following setting. Choose an American household at random and let the random variable \(X\) be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: \begin{tabular}{lcccccc} \hline Number of cars \(\mathrm{X}:\) & 0 & 1 & 2 & 3 & 4 & 5 \\ Probability: & 0.09 & 0.36 & 0.35 & 0.13 & 0.05 & 0.02 \\ \hline \end{tabular} The standard deviation of \(X\) is \(\sigma_{X}=1.08\). If many households were selected at random, which of the following would be the best interpretation of the value \(1.08 ?\) (a) The mean number of cars would be about 1.08 . (b) The number of cars would typically be about 1.08 from the mean. (c) The number of cars would be at most 1.08 from the mean.

Aircraft engines Engineers define reliability as the probability that an item will perform its function under specific conditions for a specific period of time. A certain model of aircraft engine is designed so that each engine has probability 0.999 of performing properly for an hour of flight. Company engineers test an \(\mathrm{SRS}\) of 350 engines of this model. Let \(\bar{X}=\) the number that operate for an hour without failure. (a) Explain why \(X\) is a binomial random variable. (b) Find the mean and standard deviation of \(X .\) Interpret each value in context. (c) Two engines failed the test. Are you convinced that this model of engine is less reliable than it's supposed to be? Compute \(P(X \leq 348)\) and use the result to justify your answer.

Pregnancy length The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random. Let \(X\) \(=\) the length of her pregnancy. (a) Find \(P(X \geq 240)\) and interpret the result. (b) What is \(P(X>240)\) ? Explain. (c) Find the value of \(c\) such that \(P(X \geq c)=0.20 .\) Show your work.

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