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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) 1.3935 cm, (c) 0.9544

Step by step solution

01

Define the Given Information

We are given a normal distribution representing the thickness of an aluminum cylinder. The mean thickness is \(\mu = 1.41\, \text{cm}\) and the standard deviation is \(\sigma = 0.01\, \text{cm}\). We need to solve three separate parts regarding probabilities and thresholds for this normal distribution.
02

Calculate Probability for Thickness Greater than 1.42 cm

To find the probability that a thickness is greater than \(1.42\, \text{cm}\), we first calculate the Z-score using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X = 1.42\). This yields \(Z = \frac{1.42 - 1.41}{0.01} = 1\). We then find the corresponding probability from the standard normal distribution table, which tells us the probability that \(Z\) is less than 1. The probability that \(Z > 1\) is \(1 - P(Z < 1)\), or \(1 - 0.8413 = 0.1587\).
03

Find Thickness Exceeded by 95% of Samples

Since we are looking for the thickness exceeded by 95% of the samples, we actually need the point below which 5% of the samples lie (the 5th percentile). For a normal distribution, we find the Z-score corresponding to 0.05, which is approximately \(-1.645\). Then, we use the Z-score formula in reverse: \(X = \mu + Z \cdot \sigma = 1.41 + (-1.645) \cdot 0.01 = 1.3935\). Therefore, 95% of the samples exceed a thickness of \(1.3935\, \text{cm}\).
04

Calculate Proportion of Samples Meeting Specifications

To find the proportion of samples with thickness between \(1.39\, \text{cm}\) and \(1.43\, \text{cm}\), we calculate two Z-scores: \(Z_{1.39} = \frac{1.39 - 1.41}{0.01} = -2\) and \(Z_{1.43} = \frac{1.43 - 1.41}{0.01} = 2\). Using the standard normal distribution, we find \(P(Z < -2) = 0.0228\) and \(P(Z < 2) = 0.9772\). The probability of thickness between \(1.39\, \text{cm}\) and \(1.43\, \text{cm}\) is \(P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544\).
05

Summarize and Conclude

We have: (a) The probability that thickness is greater than \(1.42\, \text{cm}\) is \(0.1587\). (b) The thickness exceeded by 95% of the samples is \(1.3935\, \text{cm}\). (c) The proportion of samples meeting the specification \([1.39, 1.43]\, \text{cm}\) is \(0.9544\).

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Key Concepts

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

Probability Calculation
Probability in the context of a normal distribution is a measure of how likely a given event, or range of outcomes, is to happen. To calculate probabilities for a normally distributed variable, we often use Z-scores. These scores help us transform our normal distribution into a standard normal distribution, which has a mean of 0 and a standard deviation of 1. By converting our original values into Z-scores, we can use standard normal distribution tables to find probabilities.

For example, when calculating the probability that the thickness is greater than 1.42 cm, we first calculate the Z-score for 1.42 cm. This translation into the Z-score format allows us to utilize these tables effectively. Once the Z-score is determined, finding the probability that a value is greater than a specific Z-score is simply a matter of subtracting the cumulative probability from 1. This process highlights the power and utility of mapping real-world measureables onto a well-understood statistical framework.
Standard Deviation
Standard deviation is a fundamental concept in statistics that measures how much individual data points differ from the mean of a dataset. In simpler terms, it indicates the "spread" or "dispersion" of a dataset relative to its mean. For a normal distribution, which is symmetric about the mean, the standard deviation plays an especially critical role as it allows us to understand the distribution's shape and behavior.

Consider the aluminum cylinder's thickness distribution with a standard deviation of 0.01 cm. This small standard deviation tells us that most thickness measurements should be close to the mean, 1.41 cm. It's important because it impacts calculations that determine probabilities or specific thresholds. In a more practical sense, this small variation means our manufacturing process is quite consistent, which is often desirable in quality control scenarios.
Z-score
A Z-score is a numerical measurement that describes how many standard deviations a specific data point is from the mean of the data set. The formula to calculate a Z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

The Z-score offers a standardized way to consider how extreme or typical a particular measure is relative to the whole group. In this exercise, calculating Z-scores helps determine probabilities and specific points in the distribution. For example, we used Z-scores to find probabilities for a thickness greater than 1.42 cm and to identify the thickness exceeded by a certain percentage of samples.
Z-scores are particularly beneficial because they allow comparisons across different datasets or even different units, standardizing the data regardless of the original scale.
Statistical Analysis
Statistical analysis is the process of collecting, examining, and interpreting data to uncover patterns and trends. With the normal distribution as a backdrop, it becomes a powerful tool for insight, prediction, and making informed decisions.

In this exercise, statistical analysis enables us to understand the probability distributions of thickness, assess what thickness levels are typical, and evaluate how many samples meet given specifications. By applying methods such as probability calculations and Z-score evaluations, we extract meaningful conclusions from data that otherwise might be inscrutable.

For practical applications, this allows an organization to optimize processes or check if certain specifications will generally be met based on past performance data. Therefore, statistical analysis doesn't merely answer questions about current datasets but also informs future practice and decision-making.

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