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Components coming off an assembly line are either free of defects (S, for success) or defective (F, for failure). Suppose that \(70 \%\) of all such components are defect-free. Components are independently selected and tested one by one. Let \(y\) denote the number of components that must be tested until a defect-free component is obtained. a. What is the smallest possible \(y\) value, and what experimental outcome gives this \(y\) value? What is the second smallest \(y\) value, and what outcome gives rise to it? b. What is the set of all possible \(y\) values? c. Determine the probability of each of the five smallest \(y\) values. You should see a pattern that leads to a simple formula for \(p(y)\), the probability distribution of \(y\).

Short Answer

Expert verified
a. The smallest possible \(y\) value is 1 if the first component tested is defect-free; The second smallest \(y\) value is 2 if the first component tested is defective and the second one is defect-free. \n b. The set of all possible \(y\) values is all positive integers, \(y \in \{1, 2, 3, \ldots\}\). \n c. The probabilities for the first five \(y\) values are approximately 0.7, 0.21, 0.063, 0.019, 0.006, following a geometric distribution.

Step by step solution

01

Determine Smallest \(y\) Value

The smallest possible value of \(y\) is 1. This would occur if the very first component tested is defect-free, meaning that only one component needed to be tested until a defect-free component was obtained.
02

Identify Second Smallest \(y\) Value

The second smallest possible value of \(y\) is 2. This would happen if the first component tested is defective and the second one is defect-free. The value of \(y\) is 2 because it took two components tested until a defect-free component was found.
03

Identify the Set of All Possible \(y\) Values

Since \(y\) is the number of components needed to be tested until a defect-free component is obtained, and it's theoretically possible that infinite many components could be defective, the set of all possible \(y\) values would be the set of all positive integers, \(y \in \{1, 2, 3, \ldots\}\).
04

Calculate Probability for Each \(y\) Value

A geometric distribution can be calculated using the formula \(p(y) = (1-p)^(y-1)*p\), where \(p\) is the probability of success. Given that the probability of a component being defect-free (success) is \(p = 0.7\), the probabilities of each \(y\) value are as follows: For y = 1, \(p(y=1) = (1-0.7)^(1-1)*0.7 = 0.7\)For y = 2, \(p(y=2) = (1-0.7)^(2-1)*0.7 \approx 0.21\)For y = 3, \(p(y=3) = (1-0.7)^(3-1)*0.7 \approx 0.063\)For y = 4, \(p(y=4) = (1-0.7)^(4-1)*0.7 \approx 0.019\)For y = 5, \(p(y=5) = (1-0.7)^(5-1)*0.7 \approx 0.006\)

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Key Concepts

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

Probability Distribution
Understanding the probability distribution is crucial in statistical theory, as it helps us characterize the likelihood of different outcomes. In our context, we focus on the geometric distribution, a type of discrete probability distribution that models the number of trials needed to achieve the first success. This distribution applies to scenarios where each trial is independent and has only two possible outcomes: success or failure.

The geometric distribution is defined by the probability of success in each individual trial, denoted as 'p'. If the probability of obtaining a defect-free component is 70%, we express this as 'p = 0.7'. Using the geometric distribution, we can calculate the probability of finding the first defect-free component on the first test, second test, and so on. These calculations conform to a distinct and predictable pattern, allowing us to derive a simple formula.
Defect-Free Components
In the manufacturing industry, 'defect-free components' are those that meet the desired quality standards without any flaws or issues. The pursuit of defect-free components is critical as it ensures reliability and customer satisfaction. In the given exercise, the factor of paramount importance is the probability that a randomly selected component from the assembly line is defect-free, which is given as 70%.

This notion serves as the 'success' in our probability model, contributing to our understanding of the geometric distribution. By testing components sequentially until we encounter a defect-free one, we can observe the application of this statistical concept in a tangible setting.
Probability of Success
The 'probability of success' is a fundamental aspect of any probability distribution dealing with binary outcomes, such as the geometric distribution in our case. When considering a series of independent and identical trials, as in our exercise where components are tested one after another, the probability of success is the chance that a single trial results in a defect-free component, which is stated as 70% or 0.7.

This value directly influences the shape and behavior of the probability distribution. For instance, higher probabilities of success lead to a greater likelihood that a success will occur earlier in the testing process. Conversely, lower probabilities make the search for the first success more prolonged.
Statistical Theory
Statistical theory provides the foundation for making inferences based on data collected from experiments or observations. In the context of our exercise, statistical theory underpins the concepts of probability distributions and helps us quantify the process of finding defect-free components. The geometric distribution is steeped in statistical theory as it's based on principles of randomness, independence of trials, and binary outcomes.

As such, statistical theory is not only about calculating probabilities; it also involves understanding the assumptions behind a given model and the real-world implications of these assumptions. By appreciating the role of statistical theory, we gain insight into the probability distribution's behavior, and we can predict and interpret different scenarios within the assembly line process.

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Most popular questions from this chapter

Let \(y\) denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of \(y\) is as follows: $$ \begin{array}{lrrrrl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3\), and 4 have positive probabilities. What is \(p(4)\) ? b. How would you interpret \(p(1)=.20 ?\) 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?

Two sisters, Allison and Teri, have agreed to meet between 1 and 6 P.M. on a particular day. In fact, Allison is equally likely to arrive at exactly 1 P.M., 2 P.M., 3 P.M., 4 P.M., 5 P.M., or 6 P.M. Teri is also equally likely to arrive at each of these six times, and Allison's and Teri's arrival times are independent of one another. Thus there are 36 equally likely (Allison, Teri) arrival-time pairs, for example, \((2,3)\) or \((6,1)\). Suppose that the first person to arrive waits until the second person arrives; let \(w\) be the amount of time the first person has to wait. a. What is the probability distribution of \(w\) ? b. How much time do you expect to elapse between the two arrivals?

The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of \(5 \mathrm{ft}\) 7 in. The suit claimed that this restriction eliminated more than \(94 \%\) of adult females from consideration. Let \(x\) represent the height of a randomly selected adult woman. Suppose that \(x\) is approximately normally distributed with mean 66 in. (5 ft 6 in.) and standard deviation 2 in. a. Is the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}\) 7 in. correct? b. What proportion of adult women would be excluded from employment as a result of the height restriction?

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the variable \(x\) as the number of people who actually show up for a sold-out flight. From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{lrrrrrrrr} x & 95 & 96 & 97 & 98 & 99 & 100 & 101 & 102 \\ p(x) & .05 & .10 & .12 & .14 & .24 & .17 & .06 & .04 \\ x & 103 & 104 & 105 & 106 & 107 & 108 & 109 & 110 \\ p(x) & .03 & .02 & .01 & .005 & .005 & .005 & .0037 & .0013 \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?

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