/*! 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 8 : Syringes The quality-control i... [FREE SOLUTION] | 91Ó°ÊÓ

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: Syringes The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose the batch contains \(1 \%\) defective syringes. (a) Make a histogram showing the probabilities of \(r=0,1,2,3,4,5,6,7\), and 8 defective syringes in a random sample of eight syringes. (b) Find \(\mu .\) What is the expected number of defective syringes the inspector will find? (c) What is the probability that the batch will be accepted? (d) Find \(\sigma\).

Short Answer

Expert verified
The expected number is 0.08. The batch is accepted with probability \( P(0) + P(1) \). The standard deviation is \( \sqrt{8 \cdot 0.01 \cdot 0.99} \).

Step by step solution

01

Define the Probability Mass Function

Since the batch contains 1% defective syringes, we model the situation using a binomial distribution with parameters \( n = 8 \) (number of trials) and \( p = 0.01 \) (probability of success, i.e., finding a defective syringe). The probability of finding exactly \( r \) defective syringes is given by the binomial formula:\[P(r) = \binom{n}{r} p^r (1-p)^{n-r}\]where \( \binom{n}{r} \) is the binomial coefficient, \( n \) choose \( r \).
02

Calculate Probabilities for Each r

Using the formula from Step 1, calculate the probability for each possible value of \( r \) from 0 to 8:- \( r = 0 \): \( P(0) = \binom{8}{0} (0.01)^0 (0.99)^8 \)- \( r = 1 \): \( P(1) = \binom{8}{1} (0.01)^1 (0.99)^7 \)- \( r = 2 \): \( P(2) = \binom{8}{2} (0.01)^2 (0.99)^6 \)- Repeat similarly for \( r = 3 \) to \( r = 8 \).Calculate these values using a calculator.
03

Create a Histogram

Draw a histogram with \( r \) values on the x-axis (0 to 8) and corresponding probabilities on the y-axis. Plot bars for each \( r \), where the height of each bar represents the probability calculated in Step 2.
04

Calculate the Mean \( \mu \)

The mean number of defective syringes, \( \mu \), from a binomial distribution is given by:\[ \mu = n \cdot p = 8 \cdot 0.01 = 0.08 \]So, \( \mu = 0.08 \).
05

Determine the Probability of Acceptance

The batch is accepted if none or only one defective syringe is found. Thus, the probability of acceptance is:\[ P(\text{accept}) = P(0) + P(1) \]Using the probabilities calculated in Step 2, find this sum.
06

Calculate the Standard Deviation \( \sigma \)

The standard deviation of a binomial distribution is calculated using the formula:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{8 \cdot 0.01 \cdot 0.99} \]Compute this value to find the standard deviation.

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Key Concepts

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

Quality Control
Quality control is a vital aspect in manufacturing processes, especially when dealing with large batches of products like syringes. The goal is to ensure that the products meet certain standards and are safe for use. In the given exercise, the quality control inspector examines a sample of syringes to determine if the entire batch should be rejected. This decision hinges on the number of defective syringes found in the sample. If the inspector finds two or more defective syringes in the sample of eight, the entire batch is rejected. This approach helps in maintaining high-quality standards, as detecting defects early can prevent faulty products from reaching consumers. Employing statistical methods like the binomial distribution is essential in making informed quality control decisions, allowing manufacturers to adhere to acceptable defect levels while minimizing waste.
Defective Rate
The defective rate is a critical parameter in quality control, as it quantifies the proportion of items expected to be defective. In this case, the defective rate is given as 1%, meaning that for every 100 syringes produced, one is expected to be faulty. Understanding the defective rate helps inspectors and manufacturers to estimate the number of defects likely to occur in a given sample, which in turn informs their acceptance or rejection decisions for the batch. When conducting statistical analyses, such as those involving binomial distributions, the defective rate is often used as the probability of 'success'—where 'success' refers to finding a defective item. Thus, in the calculation of the probability mass function, the defective rate directly influences the likelihood of finding different numbers of defective syringes in a sample.
Probability Mass Function
The probability mass function (PMF) is a fundamental concept in probability theory, particularly when dealing with discrete random variables such as the number of defective syringes in a sample. It describes the probability that a discrete random variable is exactly equal to some value. For the binomial distribution, used in our syringe quality control example, the PMF helps to determine the likelihood of observing various numbers of defective syringes in the sample. The formula used is:\[P(r) = \binom{n}{r} p^r (1-p)^{n-r}\]where \(n\) is the sample size, \(r\) is the number of defects, and \(p\) is the probability of a defect. This PMF calculation provides probabilities for observing 0 through 8 defective syringes, helping to visualize and assess the quality of the batch before deciding on acceptance or rejection.
Expected Value
The expected value, often denoted by \( \mu \), is a measure of the center of a probability distribution. For a binomial distribution, the expected value corresponds to the average number of successes (or defects, in this case) expected in a sample. It's calculated using the formula \( \mu = n \cdot p \), which in the syringe scenario, evaluates to:\[ \mu = 8 \cdot 0.01 = 0.08 \] This means that, on average, the inspector can expect to find 0.08 defective syringes in a sample of eight. Even though this value is fractional and not directly observable, it provides a theoretical benchmark around which actual observations may vary. In quality control, this expected value helps determine if there are more defects than usually anticipated, guiding the inspector to make informed acceptance or rejection decisions.
Standard Deviation
The standard deviation, represented as \( \sigma \), measures the amount of variation or dispersion in a set of values. In quality control contexts like this, it quantifies the spread of possible outcomes (i.e., the number of defective syringes) around the expected value. For a binomial distribution, the standard deviation is calculated as:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]Substituting the values for our problem, we have:\[ \sigma = \sqrt{8 \cdot 0.01 \cdot 0.99} \]This yields the standard deviation for the number of defective syringes in the sample. Understanding the standard deviation allows inspectors to gauge the reliability of the expected value. A smaller standard deviation indicates that the number of defective items is likely to be close to the expected value, while a larger standard deviation suggests wider variability. In practice, this means more consistency in inspection outcomes for tighter standard deviations.

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Most popular questions from this chapter

Sociologists say that \(90 \%\) of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). (See the source in Problem 12.) Suppose that six married women are having coffee together one morning. What is the probability that (a) all of them dislike their mother-in-law? (b) none of them dislike their mother-in-law? (c) at least four of them dislike their mother-in-law? (d) no more than three of them dislike their mother-in-law?

USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 . (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) Quota Problem How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Trevor is interested in purchasing the local hardware/sporting goods store in the small town of Dove Creek, Montana. After examining accounting records for the past several years, he found that the store has been grossing over \(\$ 850\) per day about \(60 \%\) of the business days it is open. Estimate the probability that the store will gross over \(\$ 850\) (a) at least 3 out of 5 business days. (b) at least 6 out of 10 business days. (c) fewer than 5 out of 10 business days. (d) fewer than 6 out of the next 20 business days. If this actually happened, might it shake your confidence in the statement \(p=0.60\) ? Might it make you suspect that \(p\) is less than \(0.60 ?\) Explain. (e) more than 17 out of the next 20 business days. If this actually happened, might you suspect that \(p\) is greater than \(0.60\) ? Explain.

Tailgating Do you tailgate the car in front of you? About \(35 \%\) of all drivers will tailgate before passing, thinking they can make the car in front of them go faster (Source: Bernice Kanner, Are You Normal?, St. Martin's Press). Suppose that you are driving a considerable distance on a two-lane highway and are passed by 12 vehicles. (a) Let \(r\) be the number of vehicles that tailgate before passing. Make a histogram showing the probability distribution of \(r\) for \(r=0\) through \(r=12\). (b) Compute the expected number of vehicles out of 12 that will tailgate. (c) Compute the standard deviation of this distribution.

Innocent until proven guilty? In Japanese criminal trials, about \(95 \%\) of the defendants are found guilty. In the United States, about \(60 \%\) of the defendants are found guilty in criminal trials (Source: The Book of Risks, by Larry Laudan, John Wiley and Sons). Suppose you are a news reporter following seven criminal trials. (a) If the trials were in Japan, what is the probability that all the defendants would be found guilty? What is this probability if the trials were in the United States? (b) Of the seven trials, what is the expected number of guilty verdicts in Japan? What is the expected number in the United States? What is the standard deviation in each case? (c) Quota Problem As a U.S. news reporter, how many trials \(n\) would you need to cover to be at least \(99 \%\) sure of two or more convictions? How many trials \(n\) would you need if you covered trials in Japan?

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