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Chapter 6: Problem 112

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. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

Short Answer

Expert verified
Therefore, we should recommend Supplier 2 because even though both suppliers have the same average number of flaws per glass panel (1 flaw), Supplier 2 has a smaller variance, indicating that their glass panels are more consistently made with less variability in the number of flaws.

Step by step solution

01

Calculate the Mean for Each Supplier

The mean (expected value) for a probability distribution can be calculated by multiplying each outcome by its corresponding probability, and summing up these values. For Supplier 1, the mean \(渭_x\) is calculated as follows: \(渭_x = (0*0.4) + (1*0.3) + (2*0.2) + (3*0.1) = 0 + 0.3 + 0.4 + 0.3 = 1\) For Supplier 2, the mean \(渭_y\) is calculated as: \(渭_y = (0*0.2) + (1*0.6) + (2*0.2) + (3*0) = 0 + 0.6 + 0.4 + 0 = 1\)
02

Calculate the Variance for Each Supplier

The variance (measuring the data dispersion) is given by the formula \(\variance = 螘[X^2] - (螘[X])^2\), with E[X] the expected value obtained before. For Supplier 1, we calculate \(E[X^2] = 0^2*0.4 + 1^2*0.3 + 2^2*0.2 + 3^2*0.1 = 0 + 0.3 + 0.8 + 0.9 = 2\). Then, the variance \(蟽_x^2 = E[X^2] - (螘[X])^2 = 2 - (1^2) = 1\). For Supplier 2, we calculate \(E[Y^2] = 0^2*0.2 + 1^2*0.6 + 2^2*0.2 + 3^2*0 = 0 + 0.6 + 0.8 + 0 = 1.4\). Then, the variance \(蟽_y^2 = E[Y^2] - (螘[Y])^2 = 1.4 - (1^2) = 0.4\).
03

Compare the Suppliers' Mean and Variance

Both suppliers have the same mean, but Supplier 2 has a lower variance, which means that their quality is more consistent. Worldwide, a lower variance is generally better, as it means less fluctuation in quality.

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Key Concepts

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

Understanding Expected Value
When it comes to probability distributions, the expected value is a fundamental concept that represents the average outcome if an experiment were repeated many times. Technically, it's the long-term average result of a random variable from an infinite number of trials of a stochastic process.

For instance, if you were to continuously select glass panels from a supplier, the expected value tells you the average number of flaws you're likely to encounter on those panels. To calculate the expected value, you would weigh each outcome by its probability and sum these products together. If the expected value for a supplier is high, it usually indicates that you can expect more flaws on average when you purchase their glass panels.

However, understanding the expected value alone isn't enough. It's crucial to consider it along with other statistics like variance and data dispersion to get a complete picture of what's happening within a dataset.
Variance and Quality Consistency
Variance is a statistical measure that quantifies the spread of numbers in a dataset around their mean. It strongly influences decision-making in production quality and other areas. In probability distributions, variance signifies the data dispersion or how much the random variable deviates from its expected value. A lower variance indicates that the data points are clustered closer to the mean, suggesting more consistency.

In our example, although both suppliers have the same expected value for flaws, variance sheds light on their consistency. A lower variance, as shown by Supplier 2, typically appeals to manufacturers because it signifies a more predictable quality. When the stakes are high, like in manufacturing flat-screen TVs, choosing a supplier with a lower variance might minimize risk and point towards better overall quality control.
Data Dispersion in Decision Making
Data dispersion is an overall term for the spread of data within a dataset. It encompasses measures such as range, variance, and standard deviation. Understanding data dispersion is crucial for manufacturers and businesses, as it helps gauge the reliability and predictability of outcomes.

For a manufacturer selecting a glass panel supplier, a low data dispersion would mean that the panels are mostly within the same quality range, leading to fewer surprises during production. Analyzing both the expected value, which gives the average flaw count, and the variance, which indicates how widespread those flaws are, provides a comprehensive view of the supplier's quality.

Think of it as GPS navigation: the expected value tells you your destination, while data dispersion gives you the possible routes with all their turns and detours. Less dispersion means a straighter path to your goal, which in manufacturing, equates to consistent product quality.

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Most popular questions from this chapter

The distribution of the number of items produced by an assembly line during an 8 -hour shift can be approximated by a normal distribution with mean value 150 and standard deviation 10 . a. What is the probability that the number of items produced is at most \(120 ?\) b. What is the probability that at least 125 items are produced? c. What is the probability that between 135 and 160 (inclusive) items are produced?

State whether each of the following random variables is discrete or continuous. a. The number of courses a student is enrolled in b. The time spent completing a homework assignment c. The length of a person's forearm d. The number of times out of 10 throws that a dog catches a Frisbee

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A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\). What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use
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