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Lie detectors A federal report finds that lie detector tests given to truthful persons have probability about 0.2 of suggesting that the person is deceptive." company asks 12 job applicants about thefts from previous employers, using a lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. Let \(X=\) the number of people who the lie detector says are being deceptive. (a) Find and interpret \(\mu_{X}\) . (b) Find and interpret \(\sigma_{X}\) .

Short Answer

Expert verified
(a) \(\mu_{X} = 2.4\); expected number of false positives. (b) \(\sigma_{X} \approx 1.39\); variability in false positives.

Step by step solution

01

Identify the distribution

In this exercise, we are dealing with a binomial distribution. Each job applicant takes a lie detector test, and the result is either truthful or deceptive. Since we know the probability of a lie detector being wrong (suggesting truthful persons are deceptive) is 0.2, this forms the basis of a binomial probability distribution with parameter \(p = 0.2\).
02

Determine the parameters of the binomial distribution

The total number of applicants (trials) \(n\) is 12, and the probability of suggesting deception when truthful is \(p = 0.2\). Thus, we use the binomial distribution characterized by \(n = 12\) and \(p = 0.2\).
03

Calculate the expected value \(\mu_{X}\)

The expected value \(\mu_{X}\) for a binomial distribution is given by the formula \(\mu_{X} = n \cdot p\). Substitute \(n = 12\) and \(p = 0.2\) to find \(\mu_{X} = 12 \times 0.2 = 2.4\).
04

Interpret \(\mu_{X}\)

This tells us that, on average, we expect that the lie detector will incorrectly identify \(2.4\) of the 12 truthful applicants as deceptive.
05

Calculate the standard deviation \(\sigma_{X}\)

The standard deviation \(\sigma_{X}\) for a binomial distribution is given by \(\sigma_{X} = \sqrt{n \cdot p \cdot (1-p)}\). Using \(n = 12\), \(p = 0.2\), and \(1-p = 0.8\), we find \(\sigma_{X} = \sqrt{12 \times 0.2 \times 0.8} \approx \sqrt{1.92} \approx 1.3856\).
06

Interpret \(\sigma_{X}\)

The standard deviation \(\sigma_{X}\) tells us that there is a spread of about 1.39 applicants in the number of truthful applicants who are expected to be falsely identified as deceptive by the lie detector.

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Key Concepts

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

Expected Value
When working with a binomial distribution, the expected value represents the average outcome you would anticipate over numerous trials or experiments. It's like predicting the average number of times something occurs out of a certain number of tries. In our exercise, where a company assesses 12 truthful job applicants using a lie detector, the expected value helps determine how many of these might be incorrectly labeled as deceptive. The formula for finding the expected value (\[ \mu_{X} = n \cdot p \] where - \( n \) represents the number of trials (or applicants in this case), and - \( p \) is the probability of a single event occurring (lie detector labeling a truthful person as deceptive, which is 0.2). Given the values from our scenario, with 12 applicants and a 0.2 probability of incorrect detection:- The expected value is calculated as \( \mu_{X} = 12 \times 0.2 = 2.4 \). This means, on average, we expect about 2.4 out of 12 applicants to be mistakenly identified as deceptive. While you can't have a fraction of a person, this value indicates a trend we'd see over many batches of 12 applicants.
Standard Deviation
Standard deviation gives us an idea of how much our results will vary from the expected value. It measures the spread or "dispersion" of possible outcomes. If the standard deviation is low, it means most of our results are close to the expected value. If it's high, our results are spread out over a wider range. For a binomial distribution, the standard deviation is calculated using the formula:\[ \sigma_{X} = \sqrt{n \cdot p \cdot (1-p)} \] In our situation:- \( n \) is 12 (the number of applicants), - \( p \) is 0.2 (probability of false deceptive result), and- \( 1-p \) is 0.8 (probability of a truthful outcome). Plugging these into the formula gives:\[ \sigma_{X} = \sqrt{12 \times 0.2 \times 0.8} = \sqrt{1.92} \approx 1.39 \]. This indicates that the number of applicants falsely identified as deceptive typically varies by about 1.39 from the expected 2.4. It's a measure of the reliability and consistency of the lie detector test's outcome.
Probability
Probability is the backbone of any statistical distribution, including the binomial distribution. It's a measure of how likely a particular event is to occur. In simpler terms, it tells us the chances of a specific outcome happening. In our lie detector example, we're working with a probability of 0.2. This means there is a 20% chance that the lie detector will incorrectly indicate a truthful person as deceptive. In a broader sense, - Probability allows us to compute expected values and standard deviations by offering the likelihood estimates necessary for those calculations. Understanding probability helps us make sense of the potential outcomes and guides us in interpreting the expected value and standard deviation results. By using probability, - We can predict results, - Understand uncertainties, and - Make informed decisions based on statistical models. In essence, it serves as a foundational stone for analyzing and understanding real-world problems statistically.

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