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Of the parts produced by a particular machine, \(0.5 \%\) are defective. If a random sample of 10 parts produced by this machine contains 2 or more defective parts, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on this sampling plan.

Short Answer

Expert verified
The probability of the machine being shut down is the sum of the individual probabilities of having 2 to 10 defective parts.

Step by step solution

01

Define the Binomial Distribution Parameters

A binomial distribution problem consists of the number of trials \(n\), the probability of success \(p\), and the number of success \(k\). Here, the number of trials \(n\) is 10, \(p\) is \(0.005\) (the probability of a part being defective), and \(k\) is 2 or more, which means \(k\) can be 2, 3, 4,..., up to 10.
02

Calculating Individual Probabilities

The probability for exactly \(k\) defective parts in 10 trials is given as \(P(X=k) = C(n, k) * p^k * (1-p)^{n-k}\). Substituting \(n=10\), \(p=0.005\) and \(k=2\) to \(10\), calculate the individual probabilities.
03

Summing the Probabilities

Since the event of having 2 or more defective parts includes the cases where there are 2, 3, 4,..., up to 10 defective parts, sum all the individual probabilities calculated in step 2 to find the total probability.

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Key Concepts

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

Probability Calculations
In the context of the problem, probability calculations involve determining how likely it is that a given event will happen. Here, we're interested in the probability that at least 2 out of 10 sampled parts from a machine are defective. This uses the formula for binomial distribution, which helps us understand and compute the likelihood of certain outcomes in multiple trials, given a fixed probability of success (in this case, a part being defective).

Understanding the formula: - The formula for the probability of getting exactly \(k\) successes (defective parts) in \(n\) trials is \(P(X=k) = C(n, k) * p^k * (1-p)^{n-k}\). - \(C(n, k)\) is the combination formula, representing the number of ways \(k\) successes can occur in \(n\) trials.- \(p\) is the probability of a single success on each trial. Here, it is the chance that a single part is defective, which is \(p = 0.005\).- \((1-p)\) is the probability of failure, a part not being defective.

For our specific problem, since we need to find the probability of having 2 or more defective parts, the task requires computing several probabilities \(P(X=2)\) to \(P(X=10)\), and then summing them up. This can involve a bit of number crunching since it requires the use of combinatorics and understanding exponents, but ultimately it gives a single probability that informs whether the machine should be shut down for repairs.
Statistical Sampling
Statistical sampling plays a crucial role in this exercise as it involves selecting a handful of parts (the sample) from a potentially vast production line to infer the quality of the entire process. By sampling only 10 parts, we can estimate what proportion of all parts are defective, which saves time and resources compared to examining every single part produced.

However, it's important to remember that the sample must be random and representative of the whole population. A random sample ensures that every part has an equal chance of being selected, minimizing the potential for bias.

When dealing with small sample sizes, like 10 in this case, conclusions about the population's quality become more uncertain because there's less data to base inferences upon. Larger sample sizes generally provide more reliable insights into the population characteristics. However, probability theory, like the binomial distribution used here, allows us to make informed decisions even with smaller samples, such as whether or not to perform repairs on the machine based on the sampled parts.
Defective Parts Analysis
Analyzing defective parts is critical for quality control within any manufacturing process. When manufacturers produce goods, they must ensure parts meet certain standards, and detecting defects is a key part of this effort.

In our exercise, the goal is to determine when the machine producing parts should be taken offline for maintenance or repairs. This decision hinges on the calculated probability that 2 or more parts in a sample of 10 are defective.

This analysis not only involves calculating the likelihood of defects but also understanding the implications of these findings. - If the probability is high, it indicates a greater chance of defects, justifying the machine's shutdown. - Conversely, a low probability suggests the machine operates optimally, with minimal defective output.

Overall, defective parts analysis aids in maintaining product quality, reducing waste, and preventing defective products from reaching customers, which can enhance customer satisfaction and reduce costs associated with returns or recalls.

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Most popular questions from this chapter

As a quality-control inspector of toy trucks, you have observed that \(3 \%\) of the time, the wooden wheels are bored off-center. If six wooden wheels are used on each toy truck, what is the probability that a randomly selected toy truck has no off-center wheels?

Bill has completed a 10-question multiple-choice test on which he answered 7 questions correctly. Each question had one correct answer to be chosen from five alternatives. Bill says that he answered the test by randomly guessing the answers without reading the questions or answers. a. Define the random variable \(x\) to be the number of correct answers on this test, and construct the probability distribution if the answers were obtained by random guessing. b. What is the probability that Bill guessed 7 of the 10 answers correctly? c. What is the probability that anybody can guess six or more answers correctly? d. Do you believe that Bill actually randomly guessed as he claims? Explain.

Consider a binomial experiment made up of three trials with outcomes of success, \(S,\) and failure, \(F,\) where \(P(S)=p\) and \(P(F)=q.\) a. Complete the accompanying tree diagram. Label all branches completely. b. In column (b) of the tree diagram, express the probability of each outcome represented by the branches as a product of powers of \(p\) and \(q.\) c. Let \(x\) be the random variable, the number of successes observed. In column (c), identify the value of\(x\) for each branch of the tree diagram. d. Notice that all the products in column (b) are made up of three factors and that the value of the random variable is the same as the exponent for the number\(p .\) Write the equation for the binomial probability function for this situation.

According to a National Safety Council report, up to \(78 \%\) of automobile collisions are a result of distractions such as text messaging, phoning a pal, or fumbling with the stereo. Consider a randomly selected group of 18 reported collisions. a. What is the probability that all of the collisions will be due to the distractions mentioned? b. What is the probability that 15 of the collisions will be due to the distractions mentioned?

A binomial random variable is based on \(n=20\) and \(p=0.4 .\) Find \(\Sigma x^{2} P(x).\)

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