/*! 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 150 A restaurant chain is planning t... [FREE SOLUTION] | 91Ó°ÊÓ

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A restaurant chain is planning to purchase 100 ovens from a manufacturer, provided that these ovens pass a detailed inspection. Because of high inspection costs, 5 ovens are selected at random for inspection. These 100 ovens will be purchased if at most 1 of the 5 selected ovens fails inspection. Suppose that there are 8 defective ovens in this batch of 100 ovens. Find the probability that the batch of ovens is purchased. (Note: In Chapter 5 you will learn another method to solve this problem.)

Short Answer

Expert verified
The final probability value obtained will be the explanation for the probability that the batch of ovens is purchased.

Step by step solution

01

Identify Knowns

The total number of ovens (N) is 100, the number of defective ovens (M) is 8, the number of ovens chosen for inspection (n) is 5. We need to find the probability (P) that at most 1 oven fails the inspection.
02

Using Hypergeometric Distribution Formula

The Hypergeometric distribution is given by the formula \[ P(X = k) = \frac{{C(M, k) * C(N-M, n-k)}}{{{C(N, n)}}} \] where k is the number of defective ovens in the sample, M is the total number of defective ovens, N is the total number of ovens and C denotes combination.
03

Calculate for 0 and 1 defective oven

Now, calculate the probability that 0 or 1 oven is defective i.e., \[ P(X \leq 1) \] This will be the sum of the probabilities of exactly 0 defective ovens and exactly 1 defective oven i.e., \[ P(X = 0) + P(X = 1) \] Substituting the values in the formula, \[ P = \frac{{C(8,0) * C(92,5)}}{{C(100, 5)}} + \frac{{C(8,1) * C(92,4)}}{{C(100, 5)}} \]
04

Final Calculation

Calculate the values of the combinations and input into the equation obtained in Step 3 to get the final probability.

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Key Concepts

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

Probability in Hypergeometric Distribution
Probability is the chance that a particular outcome will occur. In the context of a hypergeometric distribution, probability helps us figure out how likely we are to select a certain number of defective items from a batch. This is important when dealing with quality control problems, like the oven example.
The hypergeometric distribution is used when dealing with situations where there is no replacement, meaning the odds of an event change with each pick. For instance, once an oven is inspected, it doesn’t go back into the batch to be picked again later. This affects the probability of picking a defective oven with each selection.
  • The formula for hypergeometric distribution is \[ P(X = k) = \frac{{C(M, k) \times C(N-M, n-k)}}{{C(N, n)}} \]where:
    • N is the total number of items in the population,
    • M is the number of defective items,
    • n is the number of items picked,
    • k is the number of defective items you want to pick.
Understanding how this works lets you calculate necessary probabilities for decisions like whether or not to purchase the batch of ovens.
Sampling Without Replacement
Sampling is the process of selecting a subset of items from a larger set to make inferences about the whole set. In the problem we're discussing, the sampling is done without replacement, which is crucial to the scenario.
When sampling without replacement, once you select an item, it cannot be selected again. This is different from random sampling with replacement, where every item has the same chance to be selected again, regardless of prior selections. This means the probabilities change as each selection is made.
For instance, with the ovens, each time one is selected and inspected, the possible outcomes for subsequent selections are affected. This method allows us to better represent situations where the total number of items in the population decreases with each selection, which is often the case in real-world applications.
  • Sampling without replacement is applied using the hypergeometric distribution method.
  • It helps in accurate decision-making by considering the reduced sample size in each step.
This concept is vital in determining the probability discussed in the original problem.
Statistical Methods for Quality Control
Statistical methods provide a framework to make data-driven decisions, which can be applied heavily in quality control scenarios. The use of a hypergeometric distribution is one such statistical method.
In quality control, especially in manufacturing, it is crucial to ensure that products meet certain standards without inspecting each one, as is the case with the restaurant's oven selection process.
  • This involves using probability distributions to estimate the likelihood of certain outcomes—in this instance, determining the likelihood that 0 or 1 oven is defective out of a sample of 5.
  • The hypergeometric distribution allows businesses to make decisions based on probabilities, thus saving time and costs associated with inspecting every single item.
  • It also reduces the risk by balancing the cost of inspections with the risk of a defective product.
By understanding and applying these methods, companies can maintain high-quality standards while optimizing their processes efficiently.

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Most popular questions from this chapter

A pizza parlor has 12 different toppings available for its pizzas, and 2 of these toppings are pepperoni and anchovies. If a customer picks 2 toppings at random, find the probability that a. neither topping is anchovies b. pepperoni is one of the toppings

In a statistics class of 42 students, 28 have volunteered for community service in the past. If two students are selected at random from this class, what is the probability that both of them have volunteered for community service in the past? Draw a tree diagram for this problem.

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following a. \(P(A)=.49, \quad P(B)=.67\), and \(P(C)=.75\) b. \(P(A)=.71, \quad P(B)=.34\), and \(P(C)=.45\)

In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group. $$\begin{array}{l|llcc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over (C) } & 61 & 343 \\ \hline \end{array}$$ Suppose one person is randomly selected from these senior citizens. Find the following probabilities a. \(P(\) have been victimized or \(\mathrm{B}\) ) b. \(P(\) have never been victimized or \(\mathrm{C}\) )

Five hundred employees were selected from a city's large private companies and asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way classification table was prepared. $$\begin{array}{llc} \hline & \text { Yes } & \text { No } \\ \hline \text { Men } & 225 & 75 \\ \text { Women } & 150 & 50 \\ \hline \end{array}$$ a. Suppose one employee is selected at random from these 500 employees. Find the following probabilities. i. Probability of the intersection of events "woman" and "yes" ii. Probability of the intersection of events "no" and "man" b. Mention what other joint probabilities you can calculate for this table and then find them. You may draw a tree diagram to find these probabilities.

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