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Of the people passing through an airport metal detector, \(.5 \%\) activate it; let \(X=\) the number among a randomly selected group of 500 who activate the detector. a. What is the (approximate) pmf of \(X\) ? b. Compute \(P(X=5)\). c. Compute \(P(5 \leq X)\).

Short Answer

Expert verified
a. Approximate pmf: Poisson with \(\lambda = 2.5\). b. \(P(X=5) \approx 0.0668\). c. \(P(5 \leq X) \approx 0.2043\).

Step by step solution

01

Identify Type of Distribution

Given the problem, we see that each individual activating the detector is a Bernoulli event with probability \(p = 0.005\). The number of such activations from 500 trials follows a binomial distribution with parameters \(n = 500\) and \(p = 0.005\).
02

Use Binomial Approximation

Because \(n\) is large and \(p\) is small, we can use a Poisson approximation for the binomial distribution. The parameter \(\lambda\) for the Poisson distribution is \(\lambda = np = 500 \times 0.005 = 2.5\). Thus, \(X\) approximately follows a Poisson distribution with \(\lambda = 2.5\).
03

Determine PMF of X

For a Poisson random variable with parameter \(\lambda\), the probability mass function (pmf) is given by \(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\). Therefore, the approximate pmf of \(X\) is \(P(X=k) = \frac{e^{-2.5} 2.5^k}{k!}\).
04

Compute P(X=5)

To find \(P(X=5)\), substitute \(k = 5\) into the Poisson pmf: \[ P(X=5) = \frac{e^{-2.5} 2.5^5}{5!} \]. Calculate this to get an approximate probability. Evaluating gives \( P(X=5) \approx 0.0668 \).
05

Compute P(5 ≤ X)

To find \(P(5 \leq X)\), use the complement rule and cumulative distribution function: \[ P(5 \leq X) = 1 - P(X < 5) \]. This can be computed as \(1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4))\). Compute each using the Poisson pmf and sum them for the complement. This results in \(P(5 \leq X) \approx 0.2043\).

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Key Concepts

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

Poisson Approximation
When dealing with binomial distributions, sometimes it can become cumbersome to compute probabilities directly, especially if the sample size (\( n \)) is large and the probability of success (\( p \)) is small. This is where the Poisson approximation comes in handy.
The Poisson distribution offers a simplified way to approximate the binomial distribution when two conditions are met:
  • The number of trials (\( n \)) is large.
  • The success probability (\( p \)) is small.
Under these conditions, we set our Poisson parameter, \( \lambda \), to be equal to \( np \). This parameter represents the expected number of successes in the number of trials.
In the airport metal detector example, calculating using the binomial distribution is impractical due to \( n = 500 \) and \( p = 0.005 \). Therefore, we approximate it with a Poisson distribution where \( \lambda = 500 \times 0.005 = 2.5 \). This approximation simplifies finding the probability of events significantly.
Probability Mass Function
The Probability Mass Function (PMF) is a fundamental concept in probability theory used to describe the probabilities of different outcomes for discrete random variables.
For a Poisson distribution, the PMF is denoted as \( P(X=k) \) and given by the formula:\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where \( e \) is the base of the natural logarithm, \( \lambda \) is the average rate (\( np \) in Poisson approximation), and \( k \) represents the number of events (or successes) we are interested in. This formula gives the probability that a Poisson-distributed random variable \( X \) equals \( k \).
In the context of the metal detector, this translates to probabilities like \( P(X=5) \), which can be calculated using this formula, substituting \( k = 5 \) and \( \lambda = 2.5 \). The result reflects the probability of exactly 5 people activating the metal detector.
Bernoulli Trials
A Bernoulli trial is a basic experiment that results in a binary outcome: success or failure. Each trial is independent, adhering to its simplicity and ensuring that prior outcomes don't influence future ones. The probability of success is denoted by \( p \) while the failure is \( 1-p \).
In scenarios involving numerous Bernoulli trials, like evaluating if passengers activate an airport metal detector, the trials are discretely repeated:
  • Each passenger passing through is a separate trial.
  • "Activating the detector" is defined as the success, happening with a probability of \( 0.005 \).
Large numbers of trials with small probabilities collectively form a binomial distribution. However, when it's computationally feasible, especially with a large \( n \) as seen here, approximating the event counts with a Poisson distribution offers efficiency.
Thus, Bernoulli trials serve as the foundational building blocks, leading us towards complex distributions like binomial or their pragmatic cousin, the Poisson distribution.

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Most popular questions from this chapter

Suppose that you read through this year's issues of the \(N e w\) York Times and record each number that appears in a news article-the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be \(1,2, \ldots, 8\), or 9 . Your first thought might be that the leading digit \(X\) of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford's law: \(p(x)=P(1\) st digit is \(x)=\log _{10}(1+1 / x) x=1,2, \ldots, 9\) a. Compute the individual probabilities and compare to the corresponding discrete uniform distribution. b. Obtain the cdf of \(X\). c. Using the cdf, what is the probability that the leading digit is at most 3 ? At least 5 ? [Note: Benford's law is the basis for some auditing procedures used to detect fraud in financial reporting-for example, by the Internal Revenue Service.]

An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let \(p=\) the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than \(80 \%\) of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let \(X\) denote the number of homes with detectors among the 25 sampled. Consider rejecting the claim that \(p \geq .8\) if \(x \leq 15\). a. What is the probability that the claim is rejected when the actual value of \(p\) is \(.8\) ? b. What is the probability of not rejecting the claim when \(p=.7 ?\) When \(p=.6 ?\) c. How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14 ?

Let \(X\) be the damage incurred (in \$) in a certain type of accident during a given year. Possible \(X\) values are 0,1000 , 5000 , and 10000 , with probabilities .8, .1, .08, and \(.02\), respectively. A particular company offers a \(\$ 500\) deductible policy. If the company wishes its expected profit to be \(\$ 100\), what premium amount should it charge?

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha=10\) per hour. Suppose that with probability \(.5\) an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations?

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

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