/*! 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 53 Hurbert Corporation makes font c... [FREE SOLUTION] | 91Ó°ÊÓ

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Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

Short Answer

Expert verified
The exact numerical answer would depend on the computation of the probabilities in the Step 2 and Step 3, but the steps formulated would help give a clear understanding to the problem and give a numerical solution to it.

Step by step solution

01

Use binomial distribution formula

The binomial distribution probability mass function is given by: \[ P(x; n, p) = C(n, x) * p^x * (1 - p) ^{ n - x} \] Where:- \(x\) is the number of successful trials,- \(n\) is the number of total trials,- \(p\) is the probability of success on a single trial,- \(C(n, x)\) is the combination formula which calculates the number of ways you can choose \(x\) successes from \(n\) trials.
02

Calculate probability that the shipment is accepted

Given that a shipment is rejected if there are 7 or more defective cartridges from the sample of 100 cartridges, the shipment would be accepted if there are 6 or less defective cartridges. So, \(x\) varies from 0 to 6. We have \(n = 100\), \(p = 0.05\). Therefore, \[P(Accepted) = \sum_{x=0}^{6} C(100, x) * 0.05^x * (1 - 0.05)^{100 - x}\]
03

Calculate probability that the shipment is not accepted

The complement of the event of a shipment being accepted is the event that the shipment is not accepted. So the probability that the shipment is not accepted is:\[P(Not Accepted) = 1 - P(Accepted)\]

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Key Concepts

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

Probability Calculation
Probability calculation involves determining the likelihood that a particular event will occur. In the context of quality control, like in the Alpha Electronics' inspection process, this can mean understanding the odds that a shipment of cartridges will pass inspection. When calculating probability, we need to evaluate different types of events and their probabilities. For this specific problem, we calculate the probability that there are 6 or fewer defective cartridges in a sample of 100. This involves both understanding the event itself and using the correct mathematical formulas to compute the probability value. Ultimately, probability calculation allows businesses to make informed decisions about product quality and shipment acceptance.
Quality Control
Quality control is crucial for maintaining the standards of products sold and received by businesses. It involves various procedures such as inspecting products and ensuring they meet specific quality criteria. At Alpha Electronics, quality control is conducted by randomly sampling a set of cartridges from each shipment to check for defects. This process helps the company maintain its product standards and protect its brand reputation by ensuring that customers receive high-quality products. By rejecting shipments with a high percentage of defective units, Alpha Electronics minimizes the risk of customer dissatisfaction and returns. Quality control, therefore, is a vital strategy for maintaining a company's commitment to excellence.
Binomial Probability Formula
The binomial probability formula is used to find the probability of having exactly 'x' successes in 'n' independent Bernoulli trials. In our laser cartridge example, the formula is:\[ P(x; n, p) = C(n, x) \times p^x \times (1 - p)^{n - x} \]where \(C(n, x)\) is the combination function. This calculates the different ways to choose 'x' successes in 'n' trials. The formula's \'p\' represents the probability of success (in this case, finding a defective cartridge) on a single trial. "Success" in this context means finding a defective unit, even though typically, success refers to finding a desired or positive outcome.Using this formula, we calculate probabilities for different numbers of defective cartridges (from 0 to 6) to determine whether the entire shipment should be accepted.
Sample Inspection
Sample inspection is an efficient way to assess the quality of a large shipment without examining every single item. This method is especially useful in mass production settings, like in the case of Hurbert Corporation. By testing only a sample of 100 cartridges from a larger shipment, companies can save time and resources, while still gaining valuable insights about the entire batch. However, it's important to ensure that the sample is representative of the shipment. Random sampling is the best practice here, helping to provide a fair assessment of quality. Once a representative sample is chosen, it is inspected for defects. If the sample meets the set quality criteria, such as having fewer than 7 defective cartridges in our example, the entire shipment is typically deemed acceptable. This streamlined process is crucial for maintaining efficient production and shipping operations.

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Most popular questions from this chapter

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

At Jen and Perry Ice Cream Company, a machine fills 1 -pound cartons of Top Flavor ice cream. The machine can be set to dispense, on average, any amount of ice cream into these cartons. However, the machine does not put exactly the same amount of ice cream into each carton; it varies from carton to carton. It is known that the amount of ice cream put into each such carton has a normal distribution with a standard deviation of \(.18\) ounce. The quality control inspector wants to set the machine such that at least \(90 \%\) of the cartons have more than 16 ounces of ice cream. What should be the mean amount of ice cream put into these cartons by this machine?

Obtain the following probabilities for the standard normal distribution. a. \(P(z>-.98)\) b. \(P(-2.47 \leq z \leq 1.29)\) c. \(P(0 \leq z \leq 4.25)\) d. \(P(-5.36 \leq z \leq 0)\) e. \(P(z>6.07)\) f. \(P(z<-5.27)\)

Find the area under the standard normal curve a. to the right of \(z=1.36\) b. to the left of \(z=-1.97\) c. to the right of \(z=-2.05\) d. to the left of \(z=1.76\)

Let \(x\) denote the time taken to run a road race. Suppose \(x\) is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race a. in less than 160 minutes? b. in 215 to 245 minutes?

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