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The article "FBI Says Fewer than 25 Failed Polygraph Test" (San Luis Obispo Tribune, July 29,2001 ) states that false-positives in polygraph tests (i.e., tests in which an individual fails even though he or she is telling the truth) are relatively common and occur about \(15 \%\) of the time. Suppose that such a test is given to 10 trustworthy individuals. a. What is the probability that all 10 pass? b. What is the probability that more than two fail, even though all are trustworthy? c. The article indicated that 500 FBI agents were required to take a polygraph test. Consider the random variable \(x=\) number of the 500 tested who fail. If all 500 agents tested are trustworthy, what are the mean and standard deviation of \(x ?\) d. The headline indicates that fewer than 25 of the 500 agents tested failed the test. Is this a surprising result if all 500 are trustworthy? Answer based on the values of the mean and standard deviation from Part (c).

Short Answer

Expert verified
a. The probability that all 10 pass is approximately 0.19685. \nb. The probability that more than two fail, even though all are trustworthy is approximately 0.3233. \nc. With 500 agents, the mean number of agents who fail is 75, and the standard deviation is approximately 7.423. \nd. It is indeed a surprising result that fewer than 25 failed the test.

Step by step solution

01

Calculate the probability that all 10 pass

The probability that all 10 pass is given by the formula \(P(X= n)\), where \(X\) is the binomial random variable of passing the test, and \(n\) is the number of trustworthy individuals. In this case, our \(p\) is set to 0.85 (as 15% fail), and our \(n\) is 10. Therefore the probability can be calculated as \(P(X=10) = (0.85)^{10} \approx 0.19685 \).
02

Calculate the probability that more than two fail, even though all are trustworthy

This involves finding \(P(X \leq 2)\), where \(X\) is the binomial random variable of failing the test. For calculating this value, it is easier to find the sum of the probabilities of 0, 1, and 2 individuals failing and subtract this from 1. The formula is given as \(1 - [P(X=0) + P(X=1) + P(X=2)]\). After calculating, we find the probability equals to approximately 0.3233.
03

Calculate the mean and standard deviation

The formula for the mean \(µ\) of a binomial distribution is \(n*p\), and for the standard deviation \(σ\) is \(\sqrt{n*p*(1-p)}\). Substitute the given \(n=500\) and \(p=0.85\), we get \(µ =500*0.15 =75\) and \(σ = \sqrt{500*0.85*0.15} \approx 7.423\).
04

Evaluate the given situation with the calculated mean and standard deviation

The probability of failing less than 25 of 500 will be calculated as \( P(X<25)\). Also, it aids to evaluate this in terms of \(z\) scores. The \(z\) is calculated as \( \frac{x-µ}{σ}\), substituting \(x=25\), \(µ = 75\), and \(σ=7.423\) we get that the \(z \approx -6.73\) which lies beyond \(z = -3\), in the rare event area. Therefore, it is a very surprising result that fewer than 25 failed the test.

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Key Concepts

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

Understanding the Probability of Passing a Polygraph Test
The reliability of polygraph tests, commonly known as lie detector tests, is a topic of much debate. In the exercise, we deal with the probability that trustworthy individuals will pass such a test, when false positives—a scenario where the test fails a truthful person—occur 15% of the time.

Calculating the probability for one individual to pass the test is straightforward. With an 85% chance to pass (100% - 15% failure rate), for one individual, we get a probability of 0.85. When considering multiple individuals, we must account for the fact that each test is independent, meaning one person's result doesn't affect another's. To find the probability that all 10 pass, we would need to multiply the probability of one person passing the test by itself 10 times, which can be expressed mathematically as \(0.85^{10}\) and amounts to approximately 19.68%. This illustrates the compound effect of independent probabilities in binomial distributions.
Calculating Mean and Standard Deviation of a Binomial Distribution
When dealing with binomial distributions, two important statistical measures are the mean and the standard deviation. These values help us to understand the expected value and the variability of the random variable under consideration. In our case, the random variable \(X\) is the number of individuals who fail the polygraph test.

The mean of a binomial distribution is calculated as \(\mu = n\cdot p\), where \(n\) is the number of trials and \(p\) is the probability of success on a single trial. The standard deviation gives us an idea of how much variation there is from the mean and is calculated using the formula \(\sigma = \sqrt{n\cdot p\cdot (1-p)}\).

For the exercise involving 500 FBI agents, applying these formulas gives us a mean of 75 and a standard deviation of approximately 7.423. This means that, on average, we can expect 75 failures, but due to natural variability in the test results, the actual number may typically vary by about 7.423 failures from this average.
Interpreting Z-Scores
Z-scores are a standard tool for understanding how a single data point compares with a standard distribution. A z-score represents the number of standard deviations a point is from the mean. In a normal distribution context, it can tell us how unusual or common a particular outcome is.

To calculate the z-score, we use the formula \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value we are comparing to the mean (\(\mu\)) of the data, and \(\sigma\) is the standard deviation. In the FBI agents example, when we calculate a z-score for 25 failed tests out of 500, we get a z-score of approximately -6.73, which indicates that the result is 6.73 standard deviations below the mean.

Typically, a z-score beyond -3 or +3 is considered very unusual, as it falls outside the range where we expect 99.7% of all data points to lie in a normal distribution. Hence, getting a result of fewer than 25 failures is not just surprising; it is extraordinarily rare given the expected distribution, suggesting factors outside of randomness or questioning the initial assumption that all agents were, in fact, trustworthy.

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