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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 2 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 can be calculated using the Binomial Distribution formula. b. The probability that more than 2 fail can also be calculated using the Binomial Distribution formula. c. The mean and standard deviation of the number of agents failing the test can be calculated using informations about statistical properties of binomial distribution. d. If the calculated mean value is higher than 25, then it would be surprising that fewer than 25 agents failed the test as it's lower than the expected failure rate.

Step by step solution

01

Calculate probability all 10 pass

Use the binomial distribution formula to calculate the probability that all 10 pass. The formula is \(P(x) = C(n, x) * (p^x) * ((1-p)^(n-x))\). Here, x is the number of successes (people passing), n is the number of attempts (people taking the test), and p is the probability of success (chance to pass the test). The calculation should look like this: \(P(10) = C(10, 10) * (0.85^10) * ((1-0.85)^(10-10))\).
02

Calculate probability more than 2 fail

To find the probability that more than 2 fail, first find the probability that 0, 1, or 2 people fail and then subtract these from 1. Calculate \(P(0)\), \(P(1)\), and \(P(2)\) using the binomial distribution formula and then subtract the sum of these probabilities from 1. The calculations should look like this: \(P(more than 2 fail) = 1 - [P(0) + P(1) + P(2)]\).
03

Calculate the mean and standard deviation of x

In a binomial distribution, the mean \(\mu\) is \(n*p\) and the standard deviation \(σ\) is \(\sqrt{n*p*(1-p)}\). Calculate these values using n=500 and p=0.15. The calculations should look like this: \(\mu = 500*0.15\) and \(σ = \sqrt{500*0.15*(1-0.15)}\).
04

Analyze whether fewer than 25 failing is surprising

Since the mean (expected value) is calculated in Step 3, compare it with 25 to analyze the expected result. If 25 is below the calculated mean value, that means it's lower than expected and thus surprising.

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Key Concepts

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

Probability
When we talk about probability, we think about the chance of an event happening. In the context of binomial distribution, it is about finding the probability of a certain number of successes in a fixed number of trials. The binomial distribution relies on two possible outcomes for each trial, like pass or fail.

To calculate the probability using the binomial distribution formula, you use:
  • \(P(x) = C(n, x) \times p^x \times (1-p)^{n-x}\)
  • Where \(n\) is the number of trials, \(x\) is the number of successes, and \(p\) is the probability of success on an individual trial.

For instance, if you want to find out the probability that all 10 trustworthy individuals pass a polygraph test, you would use the probability each one passes (85% chance) in the formula. This gives you a specific probability of them all passing. Similarly, to find out how many might fail, you calculate the probability for 0, 1, or 2 failures and subtract those from 1 to find more than 2 failures. This helps to cover all possible outcomes of interest.

It's essential to understand these steps to effectively use the binomial distribution for calculating probabilities in real-world scenarios.
Standard Deviation
Standard deviation provides a measure of the amount of variation or dispersion in a set of values. In a binomial distribution, the standard deviation gives insight into how much the number of successes deviates from the expected number, or mean, of successes.

The formula for calculating the standard deviation in a binomial distribution is:
  • \(σ = \sqrt{n \times p \times (1-p)}\)
  • Where \(n\) is the number of trials and \(p\) is the probability of success.

For example, if 500 trustworthy FBI agents are tested with an expected fail rate of 15%, the standard deviation helps show the spread of potential failing outcomes around the mean. A larger standard deviation means a wider spread of data points. So, in the context of the exercise, it helps to assess if seeing less than 25 failures is within a normal range or significantly different from what we might expect.

This helps in determining whether a result is surprising or expected based on usual outcomes.
Mean
The mean in a binomial distribution, often referred to as the expected value, is a way to predict the average number of successes for the given trials. It provides a baseline for what is normal or typical in a series of events.

The formula to calculate the mean in a binomial distribution is:
  • \(\mu = n \times p\)
  • Where \(n\) is the number of attempts and \(p\) is the probability of success on each attempt.

For example, when testing 500 trustworthy individuals with a test success rate of 85% (or fail rate of 15%), the mean number of failures can be calculated. With a mean of 75 failures, it acts as our chief point of comparison. If the observed number of failures, such as fewer than 25, is distinctly different from the mean, it suggests that the result might be unexpected or surprising.

Understanding how the mean works helps in setting expectations and identifying results that are significant or out of the ordinary within a particular context.

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Most popular questions from this chapter

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs [1980]: 307-325.) Let \(x\) denote the moisture loss for a randomly selected package. a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

Starting at a particular time, each car entering an intersection is observed to see whether it turns left (L) or right (R) or goes straight ahead (S). The experiment terminates as soon as a car is observed to go straight. Let \(y\) denote the number of cars observed. What are possible \(y\) values? List five different outcomes and their associated \(y\) values.

Let \(y\) denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of \(y\) is as follows: $$ \begin{array}{lrrrrl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3\), and 4 have positive probabilities. What is \(p(4)\) ? b. How would you interpret \(p(1)=.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2)\), the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(x\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

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