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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 Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example 6.23\()\)

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

A restaurant has four bottles of a certain wine in stock. The wine steward does not know that two of these bottles (Bottles 1 and 2) are bad. Suppose that two bottles are ordered, and the wine steward selects two of the four bottles at random. Consider the random variable \(x=\) the number of good bottles among these two. a. One possible experimental outcome is (1,2) (Bottles 1 and 2 are selected) and another is (2,4) . List all possible outcomes. b. What is the probability of each outcome in Part (a)? c. The value of \(x\) for the (1,2) outcome is 0 (neither selected bottle is good), and \(x=1\) for the outcome (2,4) . Determine the \(x\) value for each possible outcome. Then use the probabilities in Part (b) to determine the probability distribution of \(x\). (Hint: See Example 6.5 )
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