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The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer. (a) What is the probability that a line width is greater than 0.62 micrometer? (b) What is the probability that a line width is between 0.47 and 0.63 micrometer? (c) The line width of \(90 \%\) of samples is below what value?

Short Answer

Expert verified
(a) 0.0082 (b) 0.721 (c) 0.564 micrometers

Step by step solution

01

Understanding the Normal Distribution

The problem states that the line widths are normally distributed with a mean (\( \mu \)) of 0.5 micrometers and a standard deviation (\( \sigma \)) of 0.05 micrometers. We will be using the properties of the normal distribution and the standard normal table (z-table) to find probabilities and values.
02

Calculating Probability for Part (a)

We need to find the probability that the line width is greater than 0.62 micrometers. First, calculate the z-score for 0.62 micrometers using the formula:\[z = \frac{x - \mu}{\sigma} = \frac{0.62 - 0.5}{0.05} = 2.4\] Next, use the z-table to find the probability for \(z = 2.4\). The z-table gives the probability that a value is less than a specific z-score. Subtract this probability from 1 to find the probability of being greater:\[P(X > 0.62) = 1 - P(Z < 2.4) \]From the z-table, \(P(Z < 2.4) \approx 0.9918\). Thus, the probability is:\[P(X > 0.62) \approx 1 - 0.9918 = 0.0082\]
03

Calculating Probability for Part (b)

We need to find the probability that the line width is between 0.47 and 0.63 micrometers. First, compute the z-scores:- For 0.47 micrometers:\[z = \frac{0.47 - 0.5}{0.05} = -0.6\]- For 0.63 micrometers:\[z = \frac{0.63 - 0.5}{0.05} = 2.6\]Using the z-table, find the probabilities:- \(P(Z < -0.6) \approx 0.2743\)- \(P(Z < 2.6) \approx 0.9953\)Calculate the probability of the line width being between these values:\[P(0.47 < X < 0.63) = P(Z < 2.6) - P(Z < -0.6) \approx 0.9953 - 0.2743 = 0.7210\]
04

Calculating Line Width for Part (c)

We need to find the value below which 90% of the line widths fall. This means we need the 90th percentile of the normal distribution. Looking up 0.9 in the z-table, we find that \(z \approx 1.28\). Use the z-score formula to find the micrometer value:\[x = z \cdot \sigma + \mu = 1.28 \cdot 0.05 + 0.5 = 0.564\] Thus, 90% of the line widths are below approximately 0.564 micrometers.

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Key Concepts

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

Understanding z-score
The z-score is a key concept in statistics, particularly when dealing with normal distributions. It tells us how many standard deviations an element is from the mean. In simpler terms, it's a way to measure the distance between a data point and the mean in terms of standard deviation units. This helps in comparing different data points within the same dataset or between different datasets that have different means or standard deviations. To calculate the z-score, use the formula:\[z = \frac{x - \mu}{\sigma}\]Where:- \( x \) is the value of the data point,- \( \mu \) is the mean of the data set,- \( \sigma \) is the standard deviation. For example, if we want to find the z-score of a line width of 0.62 micrometers in our given problem, where the mean is 0.5 micrometers and the standard deviation is 0.05 micrometers, we plug the values into the formula like so:\[z = \frac{0.62 - 0.5}{0.05} = 2.4\] A z-score of 2.4 tells us that the line width of 0.62 micrometers is 2.4 standard deviations above the mean.
Understanding probability in normal distribution
When dealing with normal distributions, probability helps us understand the likelihood of a random variable falling within a particular range. The normal distribution is symmetrical, and most of the data values tend to cluster around a central peak known as the mean. To calculate the probability of a certain event in a normal distribution, we often use a z-table. This table provides the probability that a standard normal random variable is less than a given value (the area to the left of a z-score on the standard normal curve). For example, if we want to find the probability that a line width exceeds 0.62 micrometers, we first determine the z-score (which we found is 2.4). Then, using a z-table, we find:- The probability that Z < 2.4 is approximately 0.9918.To find the probability of being greater than 0.62 micrometers, subtract this from 1:\[P(X > 0.62) = 1 - 0.9918 \approx 0.0082\]This indicates that there is a very low probability (about 0.82%) that the line width will be greater than 0.62 micrometers.
Understanding percentile
A percentile gives us an understanding of the relative standing of a data point within a dataset. For instance, the 90th percentile indicates that 90% of the data falls below that value. Percentiles are used to interpret data in terms of a distribution. In our example, if we want to determine the line width below which 90% of values fall, we look for the 90th percentile of a normal distribution where the mean is 0.5 micrometers and standard deviation is 0.05 micrometers.From the z-table, the z-score corresponding to the 90th percentile is about 1.28. Using the z-score formula in reverse, we calculate the specific line width:\[x = z \cdot \sigma + \mu = 1.28 \cdot 0.05 + 0.5 = 0.564\]Thus, the 90th percentile, or the line width below which 90% of the samples fall, is approximately 0.564 micrometers.
Understanding standard deviation in normal distribution
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In a normal distribution, it helps define how spread out the data is around the mean. A smaller standard deviation means the data points are clustered closely around the mean, while a larger standard deviation indicates they are spread out over a wider range of values. This correlation between standard deviation and data distribution can provide insights into the dataset's variability. In our example, the standard deviation is 0.05 micrometers. This informs us about the average distance of each line width measurement from the mean (0.5 micrometers). It's crucial when calculating probabilities involving the normal distribution, as seen in the computation of z-scores. Understanding standard deviation is essential because: - It's a foundational component for calculating z-scores, which in turn help assess probabilities and percentiles. - It helps in comparing variability between different datasets. Standard deviation is integral for interpreting the broader trends of a dataset, helping to predict the likelihood of observing various outcomes.

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