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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

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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 \(μ_x\) is calculated as follows: \(μ_x = (0*0.4) + (1*0.3) + (2*0.2) + (3*0.1) = 0 + 0.3 + 0.4 + 0.3 = 1\) For Supplier 2, the mean \(μ_y\) is calculated as: \(μ_y = (0*0.2) + (1*0.6) + (2*0.2) + (3*0) = 0 + 0.6 + 0.4 + 0 = 1\)
02

Calculate the Variance for Each Supplier

The variance (measuring the data dispersion) is given by the formula \(\variance = Ε[X^2] - (Ε[X])^2\), 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 \(σ_x^2 = E[X^2] - (Ε[X])^2 = 2 - (1^2) = 1\). 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 \(σ_y^2 = E[Y^2] - (Ε[Y])^2 = 1.4 - (1^2) = 0.4\).
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

Suppose \(x=\) the number of courses a randomly selected student at a certain university is taking. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & 0.02 & 0.03 & 0.09 & 0.25 & 0.40 & 0.16 & 0.05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? More than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\). Explain in words why these two probabilities are different.

Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. (Hint: See Example 6.32 ) a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109\).) d. \(P(75 \leq x \leq 125)\)

A pizza shop sells pizzas in four different sizes. The 1,000 most recent orders for a single pizza resulted in the following proportions for the various sizes: $$ \begin{array}{lcccc} \text { Size } & 12 \text { in. } & 14 \text { in. } & 16 \text { in. } & 18 \text { in. } \\ \text { Proportion } & 0.20 & 0.25 & 0.50 & 0.05 \end{array} $$ With \(x=\) the size of a pizza in a single-pizza order, the given table is an approximation to the population distribution of \(x\). a. Write a few sentences describing what you would expect to see for pizza sizes over a long sequence of single-pizza orders. b. What is the approximate value of \(P(x<16)\) ? c. What is the approximate value of \(P(x \leq 16) ?\)

Seventy percent of the bicycles sold by a certain store are mountain bikes. Among 100 randomly selected bike purchases, what is the approximate probability that a. At most 75 are mountain bikes? (Hint: See Example 6.33 ) b. Between 60 and 75 (inclusive) are mountain bikes? c. More than 80 are mountain bikes? d. At most 30 are not mountain bikes?

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) 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?

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