/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Psychology: Deceit Aldrich Ames ... [FREE SOLUTION] | 91Ó°ÊÓ

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Psychology: Deceit Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies! Lies!! Lies!!! The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

Short Answer

Expert verified
(a) 0.198, (b) 0.9996, (c) 0.0004, (d) 0.000000384

Step by step solution

01

Identify the Probability of Passing

From the given information, we know that the probability of a student passing the polygraph test, given training, is 85%, or 0.85. Therefore, the probability of failing the test is 1 minus the probability of passing, which is 0.15.
02

Determine the Distribution

Since there are only two outcomes (pass or fail) and we have a fixed number of trials (9 students), this scenario follows a binomial distribution. We use this distribution to calculate the probabilities of different numbers of students passing the test.
03

Probability That All Students Pass

Using the binomial probability formula, \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \), this is the probability that all students (k=9) pass:\[ P(9 \, \text{pass}) = \binom{9}{9} (0.85)^9 (0.15)^0 = (0.85)^9 \approx 0.198 \]
04

Probability That More Than Half Pass

More than half means 5, 6, 7, 8, or 9 students pass. We calculate the cumulative probability:\[ P(X>4) = P(5) + P(6) + P(7) + P(8) + P(9) \]This can be simplified using cumulative binomial probabilities from a calculator or software, resulting in a probability of approximately 0.9996.
05

Probability That No More Than Four Pass

No more than four means 0 to 4 students pass. We calculate the cumulative probability:\[ P(0\leq X\leq4) = 1 - P(X>4) = 1 - 0.9996 = 0.0004 \]
06

Probability That All Students Fail

For all students to fail, none pass, so k=0. Using the binomial formula,\[ P(0\, \text{pass}) = \binom{9}{0} (0.85)^0 (0.15)^9 = (0.15)^9 \approx 0.000000384 \]

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Key Concepts

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

Binomial distribution
The binomial distribution is a common way to analyze experiments involving two possible outcomes—for example, pass or fail. Each student taking the polygraph test represents an independent "trial" with two potential outcomes. This makes the binomial distribution a perfect fit for calculating the probabilities in our psychology exercise.

When using a binomial distribution, there are a few important parameters to understand:
  • Number of trials (\(n\)): This refers to the total number of subjects involved, which in our case is 9 students.
  • Probability of success (\(p\)): In our exercise, success means passing the polygraph, which has been estimated at 85% or 0.85 after undergoing specific training.
  • Number of successes (\(k\)): This is the exact number of students we are calculating the probability for.
For example, the formula for calculating the probability of all 9 students passing is employing the binomial formula:\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]This formula allows a precise calculation of the likelihood of various outcomes in any binomial setup.
Probability calculation
Probability calculation is the mathematical method used to predict the chance of specific outcomes occurring. For the task concerning the polygraph test, probabilities are calculated based on possible results of students passing or failing.

The main calculations include:
  • All Students Passing: This uses the probability of success raised to the power of the number of trials (all students passing), as shown by the outcome \((0.85)^9\).
  • More Than Half the Students Passing: In this scenario, cumulative probabilities are used, adding together the probability of 5, 6, 7, 8, or 9 students passing. This can be computed using a statistical tool that accommodates binomial distributions.
  • No More Than Four Students Passing: Here, the calculation flips the previous solution by subtracting the probabilities of 5 or more students passing from 1, resulting in a much lower chance.
These calculations help us interpret the polygraph test results, emphasizing the impact of even a slight probability of passing the test after training.
Polygraph tests
Polygraph tests, often termed as lie detector tests, measure bodily responses to determine truthfulness. They are typically used in security and legal settings.

In this exercise, the polygraph test's reliability is questioned. Despite their widespread use, polygraph results can be manipulated through various countermeasures, reflecting their fallibility.
  • Measurement Techniques: Tests monitor physiological responses like heart rate, blood pressure, and breathing, assuming changes indicate deceit.
  • Counter-countermeasures: Techniques such as altered breathing or pressing toes are employed by knowledgeable individuals to manipulate results, as reflected in the exercise where training students could skew polygraph outcomes.
Understanding how easily results can be swayed underlines the importance of viewing polygraph test results with caution and skepticism, especially in high-stakes situations involving psychology or other critical fields.

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Most popular questions from this chapter

Negative Binomial Distribution: Marketing Susan is a sales representative who has a history of making a successful sale from about \(80 \%\) of her sales contacts. If she makes 12 successful sales this week, Susan will get a bonus. Let \(n\) be a random variable representing the number of contacts needed for Susan to get the 12 th sale. (a) Explain why a negative binomial distribution is appropriate for the random variable \(n\). Write out the formula for \(P(n)\) in the context of this application. Hint: See Problem 30 . (b) Compute \(P(n=12), P(n=13)\), and \(P(n=14)\). (c) What is the probability that Susan will need from 12 to 14 contacts to get the bonus? (d) What is the probability that Susan will need more than 14 contacts to get the bonus? (e) What are the expected value \(\mu\) and standard deviation \(\sigma\) of the random variable \(n\) ? Interpret these values in the context of this application.

Critical Thinking In an experiment, there are \(n\) independent trials. For each trial, there are three outcomes, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). For each trial, the probability of outcome \(\mathrm{A}\) is \(0.40 ;\) the probability of outcome \(\mathrm{B}\) is \(0.50 ;\) and the probability of outcome \(\mathrm{C}\) is \(0.10 .\) Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type \(B\), and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type \(\mathrm{A}\) and six outcomes that are not of type \(\mathrm{A}\) ? Explain. What is the probability of success on each trial?

Agriculture: Apples Approximately \(3.6 \%\) of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin (Source: Australian Journal of Agricultural Research, Vol. 25, pp. \(783-790)\). (Bitter pit is a disease of apples resulting in a soggy core, which can be caused either by overwatering the apple tree or by a calcium deficiency in the soil.) Let \(n\) be a random variable that represents the first Jonathan apple chosen at random that has bitter pit. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) Find the probabilities that \(n=3, n=5\), and \(n=12\). (c) Find the probability that \(n \geq 5\). (d) What is the expected number of apples that must be examined to find the first one with bitter pit? Hint: Use \(\mu\) for the geometric distribution and round.

Insurance: Auto The Mountain States Office of State Farm Insurance Company reports that approximately \(85 \%\) of all automobile damage liability claims are made by people under 25 years of age. A random sample of five automobile insurance liability claims is under study. (a) Make a histogram showing the probability that \(r=0\) to 5 claims are made by people under 25 years of age. (b) Find the mean and standard deviation of this probability distribution. For samples of size 5, what is the expected number of claims made by people under 25 years of age?

Defense: Radar Stations The probability that a single radar station will detect an enemy plane is \(0.65\). (a) Quota Problem How many such stations are required for \(98 \%\) certainty that an enemy plane flying over will be detected by at least one station? (b) If four stations are in use, what is the expected number of stations that will detect an enemy plane?

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