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Packaging \(\operatorname{CDs}(6.2,5.3)\) A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter \(\mu=4.2\) inches and standard deviation \(\sigma=0.05\) inches. The CD manufacturer decides to produce CDs with mean diameter \(\mu=4\) inches. Their diameters follow a Normal distribution with \(\sigma=0.1\) inches. (a) Let X the diameter of a randomly selected CD and Y the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable X Y. What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

Step by step solution

01

Define the Random Variables

Let the diameter of a randomly selected CD be represented by the random variable \(X\), where \(X\) follows a Normal distribution \(X \sim N(4, 0.1^2)\). Similarly, let the diameter of a randomly selected case be \(Y\), where \(Y \sim N(4.2, 0.05^2)\).

02

Distribution of Random Variable Z (Y - X)

The random variable of interest \(Z\), representing the clearance when a CD fits in a case, is \(Z = Y - X\). \(Y\) and \(X\) are independent normal variables, so \(Z\) is also normally distributed. The mean of \(Z\) is given by \(\mu_Z = \mu_Y - \mu_X = 4.2 - 4 = 0.2\) and the variance is \(\sigma_Z^2 = \sigma_Y^2 + \sigma_X^2 = 0.05^2 + 0.1^2 = 0.0125\). Therefore, \(Z \sim N(0.2, \sqrt{0.0125})\).

03

Importance of Z in Manufacturing

The random variable \(Z = Y - X\) measures the amount of space between the diameter of a case and a CD. A positive \(Z\) means the CD will fit inside the case. It's crucial for ensuring each CD can be packaged into a case.

04

Compute Probability of a Single CD Fitting

We need to find \(P(X < Y) = P(Y - X > 0) = P(Z > 0)\). Using the standard normal distribution, \(P(Z > 0) = P\left(\frac{Z - 0.2}{\sqrt{0.0125}} > \frac{0 - 0.2}{\sqrt{0.0125}}\right)\). Calculate the z-score: \(z = \frac{-0.2}{\sqrt{0.0125}} = -1.7889\). Look up the z-score in a standard normal distribution table to get \(P(Z > 0) \approx 0.963\) (or 96.3%).

05

Probability of All 100 CDs Fitting

Assuming independence, the probability that all 100 CDs fit is \((P(Z > 0))^{100} = 0.963^{100}\). Calculate this using a calculator or computer. \(P(all \ 100 \ fit) \approx 0.1004\).

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Key Concepts

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

Normal Distribution

In probability and statistics, the Normal Distribution plays a vital role, especially when dealing with continuous data that can be measured. It is represented by a symmetric, bell-shaped curve where most of the observations cluster around the mean. The shape of a Normal Distribution is determined by two key parameters: the mean (渭) and the standard deviation (蟽). The mean indicates the central point of the distribution, while the standard deviation measures the spread, telling us how much data tends to deviate from the average.
In practical terms, many natural phenomena follow a Normal Distribution. For example, heights, blood pressure measurements, and even the diameters of CDs and cases in our manufacturing scenario. This makes the Normal Distribution particularly valuable because it allows us to model and predict real-world situations.
In the given exercise, both the CD and case diameters follow a Normal Distribution, which facilitates calculating the probability of one fitting into the other by considering the differences between them.

Random Variables

Random variables are a foundational concept in probability, allowing us to quantify outcomes of random processes. A random variable assigns numerical values to each possible outcome of a random phenomenon. In a statistical context, it can be discrete, with specific values, or continuous, with an infinite number of possible outcomes along a continuum.
In our exercise, we deal with two continuous random variables: one for the diameter of the CDs ( X ) and another for the plastic cases ( Y ). These random variables help us model and understand the variability in the manufacturing process. Specifically, they allow us to calculate the distribution of the difference ( Z = Y - X ), which determines whether a CD fits inside a case. Understanding and working with random variables enable manufacturers to control the quality and fit of their products.

Standard Deviation

Standard deviation is a measure of how spread out numbers in a dataset are around the mean. It tells us about the variability of the data. In a Normal Distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In our manufacturing example, the standard deviation of the CD diameters is 0.1 inches, and that of the cases is 0.05 inches. These values show us how much the diameters vary from their respective average sizes. The smaller the standard deviation, the closer the data points are to the mean, indicating less variability. This is crucial for manufacturers, as it assists in ensuring consistent product quality. By understanding standard deviation, producers can predict and control deviations in their manufacturing processes more effectively.

Statistical Independence

Statistical independence means that the occurrence of one event does not affect the probability of another event occurring. In probability theory, when two variables are statistically independent, the joint probability distribution is the product of their individual distributions.
In the context of the problem, the diameters of CDs and cases ( X and Y ) are considered statistically independent. This assumption is critical because it allows us to combine the distributions of X and Y to explore their differences independently, leading to the distribution of Z = Y - X . This independence means that there is no direct relationship between the sizes of CDs and cases, permitting us to simplify calculations and predictions regarding their fit. Statistical independence is central for accurate probability computations in complex systems such as manufacturing multiple unconnected parts.