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Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

Short Answer

Expert verified
The calculated probabilities are the solutions to the exercise, which depend on the given percentage of defective parts. These probabilities are obtained by using the binomial distribution theorem and summing up the probabilities for one or fewer defective parts.

Step by step solution

01

Compute basic parameters

Firstly, identify basic parameters from the exercise. These include the number of trials \(n = 20\), and the acceptance criterium that one or fewer parts are defective. Notice that the probability of success \(p\) changes in each of the sub-problems (a, b, c). It's the percentage of defective parts in a lot.
02

Compute probabilities for sub-problem a

For sub-problem a, the defective rate is \(5 \% = 0.05\). Thus, the probability of success, which is having a defective part, is \(p = 0.05\). Use the binomial theorem to calculate the probability of accepting the lot, i.e., that the number of defective parts is one or none (zero). The binomial distribution formula is given by \[P(k; n,p) = C(n,k) \cdot (p)^k \cdot (1-p)^{n-k}\] where \(C(n, k)\) is the binomial coefficient, calculated as \[C(n,k) = n! / [k!(n-k)!]\]. Let's apply this formula for \(k = 0\) and \(k = 1\) and sum these two probabilities.
03

Compute probabilities for sub-problem b

Repeat Step 2 for sub-problem b, where the defective rate is \(10 \% = 0.10\), thus \(p = 0.10\). Use the binomial theorem as before to compute the sum of probabilities for \(k = 0\) and \(k = 1\).
04

Compute probabilities for sub-problem c

Finally, repeat Step 2 for sub-problem c, where the defective rate is \(20 \% = 0.20\), thus \(p = 0.20\). Use the binomial theorem as before to compute the sum of probabilities for \(k = 0\) and \(k = 1\).

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Key Concepts

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

Quality Control Statistics
Quality control statistics are fundamental in manufacturing and production, ensuring that products meet certain standards of quality. In the context of industrial quality control, statistical methods are used to decide whether a batch of components is satisfactory based on a sample inspection. For instance, a company might set a rule that a batch of products is acceptable if only a small proportion of sampled items are defective. This method helps in conserving resources because inspecting every single item may not be practical or cost-effective.

Statistical techniques, such as the binomial distribution, are particularly useful when dealing with yes-or-no type attributes, such as 'defective' or 'non-defective'. In practice, inspectors would randomly select a number of parts to test from a large lot. If the number of defective parts is below a predetermined threshold, the lot passes the quality check. Otherwise, it may be rejected or subjected to further inspection.
Binomial Theorem
The binomial theorem is a crucial principle in probability theory that explains the distribution of outcomes for a given number of independent experiments, or 'trials'. When the outcome of each trial has only two possible states (such as pass or fail, defective or non-defective), the probability of a specific number of successes can be calculated using the binomial distribution formula.

Incorporating the binomial theorem in quality control allows us to precisely calculate the probability of observing a certain number of defective parts in a sample. By understanding this theorem, we can find the likelihood that, given the proportion of defective items in a lot, a randomly selected sample of parts will have a satisfactory number of defects (usually set to a maximum allowable limit).
Defective Parts Inspection
Defective parts inspection is a process used to identify and segregate defective items from a production batch, ensuring that only quality products reach the end consumer. The process typically involves randomly selecting a set number of items from a lot and checking them for defects.

When implementing a sampling plan, the objective often is to balance the risk of accepting defective lots (producer's risk) against the risk of rejecting good lots (consumer's risk). In the exercise given, the lot is considered acceptable if at most one of the sampled parts is defective. This principle protects the consumer by setting strict acceptance criteria, thus ensuring high-quality products.
Binomial Distribution Formula
The binomial distribution formula provides a way to calculate the probability of obtaining a certain number of successes in a fixed number of independent trials, each with the same probability of success. The formula is expressed as:
\[P(k; n,p) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here, \(P(k; n,p)\) represents the probability of getting exactly \(k\) successes in \(n\) trials, and \(p\) is the probability of success on a single trial. The term \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes from \(n\) trials.

In the quality control scenario, this formula helps in determining the probability that a lot will be accepted or rejected, based on the defect rate and the inspection criteria. Using this formula, we can calculate the probability for a range of defective rates, assisting quality control professionals in making informed decisions about their manufacturing processes.

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Most popular questions from this chapter

A soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has 10 cans of each type. If 15 customers want to purchase drinks before the machine is restocked, what is the probability that each of the 15 is able to purchase the type of drink desired? (Hint: Let \(x\) denote the number among the 15 who want a diet drink. For which possible values of \(x\) is everyone satisfied?)

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) 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?

Example 6.14 gave the probability distributions shown below for \(x=\) number of flaws in a randomly selected glass panel from Supplier 1 \(y=\) number of flaws in a randomly selected glass panel from Supplier 2 for two suppliers of glass used in the manufacture of flat screen TVs. If the manufacturer wanted to select a single supplier for glass panels, which of these two suppliers would you recommend? Justify your choice based on consideration of both center and variability. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

The Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all tudents to complete the exam

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