/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Thickness measurements of a coat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 66

Thickness measurements of a coating process are made to the nearest hundredth of a millimeter. The thickness measurements are uniformly distributed with values 0.15 , \(0.16,0.17,0.18,\) and \(0.19 .\) Determine the mean and variance of the coating thickness for this process.

Short Answer

Expert verified
Mean: 0.17 mm, Variance: 0.0002 mm虏.

Step by step solution

01

Understand Uniform Distribution

When thickness measurements are uniformly distributed over discrete values, it means each value is equally likely to occur. In this case, values are 0.15, 0.16, 0.17, 0.18, and 0.19 millimeters.
02

Calculate the Mean

To find the mean of a uniformly distributed dataset, sum all possible values and divide by the number of values. \[\text{Mean} = \frac{0.15 + 0.16 + 0.17 + 0.18 + 0.19}{5}\]Calculating, the mean is 0.17 millimeters.
03

Calculate Variance

Variance is calculated using the formula: \[\text{Variance} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \text{Mean})^2\]Substitute the given data points:\[\text{Variance} = \frac{1}{5} [(0.15 - 0.17)^2 + (0.16 - 0.17)^2 + (0.17 - 0.17)^2 + (0.18 - 0.17)^2 + (0.19 - 0.17)^2]\]This results in:\[\text{Variance} = \frac{1}{5} [0.0004 + 0.0001 + 0.0000 + 0.0001 + 0.0004]\]Finally, calculate:\[\text{Variance} = \frac{0.001}{5} = 0.0002\] millimeters squared.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Mean Calculation
The mean is a central concept in statistics, representing the average value in a dataset. To calculate it for a set of uniformly distributed values, you add up all values and divide by the number of values. This acts like finding the balance point or center among measurements. In our case of coating thickness measurements with values 0.15, 0.16, 0.17, 0.18, and 0.19 millimeters:
  • Add all the values together: 0.15 + 0.16 + 0.17 + 0.18 + 0.19 = 0.85
  • Count the number of values, which is 5.
  • Divide the sum by the number of values: \[\text{Mean} = \frac{0.85}{5} = 0.17\]
This simple calculation helps summarize and understand the general thickness of the coating. So, the mean coating thickness is 0.17 millimeters.
Variance Calculation
Variance provides insight into the variability or spread of values from the mean in a dataset. It quantifies how much the measurements differ from the average. To calculate it:
  • First, find the mean. For our values, it's 0.17, as previously calculated.
  • Subtract the mean from each value, then square the result to eliminate negative differences:
    • \((0.15 - 0.17)^2 = 0.0004\)
    • \((0.16 - 0.17)^2 = 0.0001\)
    • \((0.17 - 0.17)^2 = 0\)
    • \((0.18 - 0.17)^2 = 0.0001\)
    • \((0.19 - 0.17)^2 = 0.0004\)
  • Sum these squared differences: \[0.0004 + 0.0001 + 0 + 0.0001 + 0.0004 = 0.001\]
  • Divide by the number of data points. Since we have 5, the variance is:\[\text{Variance} = \frac{0.001}{5} = 0.0002\]
The variance equals 0.0002 millimeters squared. This small value indicates that the coating measurements are closely packed around the mean, showing consistency in the process.
Coating Thickness Measurement
Coating thickness measurements ensure that the coating adheres to desired specifications and performance standards.
In processes where precision is critical, like in manufacturing paint finishes or protective layers, understanding the distribution of measurements helps in quality control and assurance.
  • A uniform distribution implies each measurement is equally probable, which aids in easy analysis and prediction.
  • Accurate measurements help identify variations in coating that might affect performance, such as durability or resistance.
  • By analyzing the mean and variance, professionals can quickly identify if anomalies in thickness are likely to occur and rectify them if needed.
Using these statistical tools, manufacturing processes can achieve higher quality and lower waste, ensuring coatings are applied consistently and within specifications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is one star per 16 cubic light-years. (a) What is the probability of two or more stars in 16 cubic light-years? (b) How many cubic light-years of space must be studied so that the probability of one or more stars exceeds \(0.95 ?\)

Patient response to a generic drug to control pain is scored on a 5 -point scale, where a 5 indicates complete relief. Historically, the distribution of scores is $$ \begin{array}{lllll} 1 & 2 & 3 & 4 & 5 \\ 0.05 & 0.1 & 0.2 & 0.25 & 0.4 \end{array} $$ Two patients, assumed to be independent, are each scored. (a) What is the probability mass function of the total score? (b) What is the probability mass function of the average score?

The probability that an individual recovers from an illness in a one-week time period without treatment is 0.1. Suppose that 20 independent individuals suffering from this illness are treated with a drug and four recover in a one- week time period. If the drug has no effect, what is the probability that four or more people recover in a oneweek time period?

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.02 failure per hour. (a) What is the probability that the instrument does not fail in an eight-hour shift? (b) What is the probability of at least one failure in a 24 -hour day?

Samples of 20 parts from a metal punching process are selected every hour. Typically, \(1 \%\) of the parts require rework. Let \(X\) denote the number of parts in the sample of 20 that require rework. A process problem is suspected if \(X\) exceeds its mean by more than three standard deviations. (a) If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that \(X\) exceeds its mean by more than three standard deviations? (b) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds \(1 ?\) (c) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds 1 in at least one of the next five hours of samples?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.