/*! 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 19 Criminal Justice: Jury Duty Have... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Criminal Justice: Jury Duty Have you ever tried to get out of jury duty? About \(25 \%\) of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York). If 12 people are called for jury duty, (a) what is the probability that all 12 will be available to serve on the jury? (b) what is the probability that 6 or more will not be available to serve on the jury? (c) Find the expected number of those available to serve on the jury. What is the standard deviation? (d) Quota Problem How many people \(n\) must the jury commissioner contact to be \(95.9 \%\) sure of finding at least 12 people who are available to serve?

Short Answer

Expert verified
(a) 0.0317; (b) 0.366; (c) Expected: 9, SD: 1.5; (d) Contact 16 people.

Step by step solution

01

Define Probabilities

Let the probability that an individual person will find an excuse be \(p = 0.25\). Therefore, the probability that a person will be available is \(q = 0.75\) since \(q = 1 - p\).
02

Calculate Probability for Part (a)

We need to find the probability that all 12 people will serve, which means none find an excuse. We use the formula for binomial probabilities: \(P(X=k) = \binom{n}{k} p^k q^{n-k}\), where \(n = 12\), \(k = 0\), and \(p = 0.25\).\[P(X = 0) = (0.75)^{12}\approx 0.0317\].
03

Calculate Probability for Part (b)

To find the probability that 6 or more people will not be available, we need to calculate\[P(X \geq 6) = 1 - P(X < 6)\]. Use binomial probabilities to find \(P(X<6)\), where \(X\) is the number not available:\[P(X < 6) = \sum_{k=0}^{5} \binom{12}{k} (0.25)^k (0.75)^{12-k}\]. After calculation, we find \[P(X \geq 6) = 1 - P(X < 6) \approx 0.366\].
04

Find Expected Number for Part (c)

The expected number \(E(X)\) of people available is given by \(np\), where \(n = 12\) and \(p = 0.75\).\[E(X) = 12 \times 0.75 = 9\].
05

Calculate Standard Deviation for Part (c)

The standard deviation \(\sigma\) in a binomial distribution is given by \(\sigma = \sqrt{npq}\). Here:\[\sigma = \sqrt{12 \times 0.75 \times 0.25} = \sqrt{2.25} \approx 1.5\].
06

Calculate for Part (d)

We want \(P(Y \geq 12) = 0.959\), where \(Y\) is the number available. Use the normal approximation method to find \(n\) so that \(P(Z \leq \frac{y - np}{\sqrt{npq}} ) \approx 0.959\). Solving iterative values or using software yields \(n \approx 16\) people need to be called.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Probability
Probability is a measure of the likelihood that a specific event will occur. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the context of jury duty, each person who is called to serve has a certain probability of either being excused or available.
The probability that a person finds an excuse and cannot serve is given as 0.25. To find the probability that someone is available, we subtract this figure from 1, resulting in 0.75. This probability helps us calculate the likelihood of different outcomes, such as how many people will or will not be available when called.
Understanding these probabilities is crucial for calculating the expected number of jurors and evaluating how many individuals may need to be contacted to ensure a sufficient jury.
Expected Value in a Binomial Distribution
Expected value (often denoted as E(X)) is a key concept in probability and statistics. It represents the average outcome of a random event if it could be repeated many times. In this scenario, we're interested in how many of the 12 people called will typically be available for jury duty.
Using the formula for expected value in the context of binomial distribution, where we have a fixed number of trials, we multiply the total number of trials () by the probability of success (here, being available for jury duty).
The expected number of people available is calculated as: \[E(X) = n \times p = 12 \times 0.75 = 9\]This means, on average, 9 out of 12 people called will be available to serve on the jury. This expected value gives us an insight into what usually happens in a similar setting.
Understanding Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the outcomes can be from the expected value.
For our jury duty example, standard deviation helps us understand the potential variability in how many people are actually available. We calculate it using the formula for standard deviation in a binomial distribution: \[\sigma = \sqrt{n \times p \times q}\]Plugging in our numbers: \[\sigma = \sqrt{12 \times 0.75 \times 0.25} = \sqrt{2.25} \approx 1.5\]This indicates that the number of people available could typically vary by about 1.5 people from the expected value (9), helping us understand the consistency of our outcome.
Calculating Binomial Probabilities
Binomial probabilities allow us to determine the likelihood of a specific number of successes in a fixed number of trials, given a constant probability of success. This is especially useful in scenarios like our jury duty problem.
Let's break down the calculation for finding specific probabilities. For instance, the probability that all 12 people are available, without any excuse, is calculated using:\[P(X = 0) = (0.75)^{12} \approx 0.0317\]Meaning there's a 3.17% chance that everyone will serve.
To find the probability of 6 or more people not being available, we first calculate the probability of less than 6 people finding excuses, then subtract it from 1 for the complement probability:\[P(X < 6) = \sum_{k=0}^{5} \binom{12}{k} (0.25)^k (0.75)^{12-k}\]Using the above sum, we compute the complement to find that about 36.6% of the time, 6 or more will be excused, demonstrating the usefulness of binomial probabilities in predicting real-world outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Critical Tbinking Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about \(6.9\). Use the criterion that it is unusual to have data values more than \(2.5\) standard deviations above the mean or \(2.5\) standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b) Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c) Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.

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?

Criminal Justice: Drunk Drivers Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 569, of which only 1 became a repeat offender. (a) Suppose that of 1000 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still \(p=1 / 569 .\) Explain why the Poisson approximation to the binomial would be a good choice for \(r=\) number of repeat offenders out of 1000 convicted drunk drivers who toured the morgue. What is \(\lambda\) to the nearest tenth? (b) What is the probability that \(r=0 ?\) (c) What is the probability that \(r>1 ?\) (d) What is the probability that \(r>2 ?\) (e) What is the probability that \(r>3 ?\)

Sociology: Ethics The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party, and then returned the purchase? This is called a one-time fling. About \(10 \%\) of all adults deliberately do a one-time fling and feel no guilt about it (Source: Are You Normal?, by Bernice Kanner, St. Martin's Press). In a group of seven adult friends, what is the probability that (a) no one has done a one-time fling? (b) at least one person has done a one-time fling? (c) no more than two people have done a one-time fling?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.