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In an accelerator center, an experiment needs a \(1.41-\mathrm{cm}-\) thick aluminum cylinder (http://puhepl.princeton.edu/mumu' target/Solenoid_Coil.pdf). Suppose that the thickness of a cylinder has a normal distribution with a mean of \(1.41 \mathrm{~cm}\) and a standard deviation of \(0.01 \mathrm{~cm} .\) a. What is the probability that a thickness is greater than \(1.42 \mathrm{~cm} ?\) b. What thickness is exceeded by \(95 \%\) of the samples? c. If the specifications require that the thickness is between \(1.39 \mathrm{~cm}\) and \(1.43 \mathrm{~cm}\), what proportion of the samples meets specifications?

Short Answer

Expert verified
a. 0.1587 b. 1.3936 cm c. 95.44%

Step by step solution

01

Understanding the Problem

The exercise provides information about the thickness of an aluminum cylinder described by a normal distribution with a mean of 1.41 cm and a standard deviation of 0.01 cm. We are asked to find probabilities and specific thickness values based on this distribution.
02

Step A - Convert to Standard Normal Variable

To find the probability that the thickness is greater than 1.42 cm, we convert this value to a standard normal variable (Z-score). The formula is: \[ Z = \frac{X - \mu}{\sigma} \]Here, \(X = 1.42\), \(\mu = 1.41\mathrm{~cm}\) and \(\sigma = 0.01\mathrm{~cm}\). Substitute these values to get:\[ Z = \frac{1.42 - 1.41}{0.01} = \frac{0.01}{0.01} = 1 \]
03

Step A - Find Probability for Z

Using a standard normal distribution table or calculator to find the probability that corresponds to \(Z = 1\). We are interested in the probability that thickness is greater than 1.42 cm:\[ P(Z > 1) = 1 - P(Z \leq 1) \approx 1 - 0.8413 = 0.1587 \]So, the probability is approximately 0.1587.
04

Step B - Determine Z-Score for 95% Exceedance

For part b, we need to find the Z-score that leaves 5% in the right tail (since \(100\% - 95\% = 5\%\) exceedance). From Z-tables, \(P(Z < z) = 0.95\) corresponds to \(z \approx -1.645\) or \(P(Z > z) = 0.05\) corresponds to \(z \approx 1.645\). We want the lower 5%, so we actually use \(-1.645\) for this problem to find the lower exceedance.
05

Step B - Convert Z-Score to Thickness

Now, use the Z-score to find the actual thickness:\[ X = \mu + Z\sigma \]\[ X = 1.41 + (-1.645)(0.01) = 1.41 - 0.01645 = 1.39355 \]Therefore, 95% of the samples have a thickness exceeding approximately 1.3936 cm.
06

Step C - Calculate Probability Within Specifications

For part c, we calculate the probability that thickness is between 1.39 cm and 1.43 cm.First, find Z-scores for both limits:For 1.39 cm: \[ Z = \frac{1.39 - 1.41}{0.01} = -2 \]For 1.43 cm: \[ Z = \frac{1.43 - 1.41}{0.01} = 2 \]Now, find probabilities:\[ P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) \approx 0.9772 - 0.0228 = 0.9544 \]So, approximately 95.44% of the samples meet the specifications.

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Key Concepts

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

Probability Calculations
Probability calculations help us determine the likelihood of an event occurring. In the context of the problem, this involves calculating the probability of finding a certain thickness within the given range of the aluminum cylinder. By utilizing the properties of the normal distribution, we can make these calculations more systematic.

To determine the probability of a cylinder being thicker than 1.42 cm, for instance, we calculate the area under the curve beyond this value. Remember these key points about probability calculations in normal distributions:
  • Create a visual mental image: think of the normal curve where the total area equals 1 (representing 100% probability).
  • Probabilities correspond to areas under this curve, allowing us to determine the fraction of occurrences above or below certain values.
  • Use of cumulative distribution functions (like Z-tables) helps translate Z-scores into probabilities.
These calculations are foundational in fields ranging from quality control processes to scientific research, where determining the likelihood of specific outcomes is crucial.
Z-score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. When the dataset follows a normal distribution, the Z-score helps standardize different variables on a single scale.

In the problem, to find how unusual or ordinary a wall thickness of 1.42 cm is, we calculate its Z-score. The formula is:\[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the thickness, \(\mu\) is the mean thickness, and \(\sigma\) is the standard deviation.

Here's how it applies:
  • Z = 1 means the thickness is one standard deviation above the mean.
  • Positive Z-scores indicate values above the mean; negative scores indicate values below.
  • It transforms data so you can easily read probabilities using the standard normal distribution table.
Understanding Z-scores is vital for measuring how far and in which direction a value deviates from the dataset's average.
Statistical Analysis
Statistical analysis is the method of understanding, interpreting, and using data to draw conclusions. It’s an umbrella term for various specific methods used, like those in the exercise to analyze the cylinder thickness data.

This problem required using statistical tools such as:
  • Normal distribution assumptions to describe the data behavior.
  • Z-scores to determine standard distances of particular values from the mean.
  • Probability calculations to find the likelihood of specific outcomes.
In the analysis, once you determine the Z-scores, you use statistical tables or calculators to find associated probabilities. This consistent and structured approach allows precise conclusions about how much of the dataset falls within specified bounds.

Each calculation and interpretation step builds upon knowledge of probability theory and normal distribution characteristics, emphasizing accurate observation and prediction.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are around the mean. It's one of the key concepts in analyzing normal distributions.

In the given exercise, the standard deviation of 0.01 cm signals the typical variation from the mean thickness of 1.41 cm. Here's how standard deviation functions within this context:
  • Small standard deviation indicates tight clustering around the mean, suggesting consistent cylinder thickness.
  • Large standard deviation would imply greater variability, possibly indicating inconsistent production.
  • By standardizing data into Z-scores, you utilize the standard deviation to express data variability in comparative terms.
Thus, comprehending standard deviation is vital for interpreting how values differ from the average and ensures you can assess consistency across samples, making it fundamental for quality assurance and process control.

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