/*! 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 1 46\. Young's modulus is a quanti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 1

46\. Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a particular type, its mean value and standard deviation are \(70 \mathrm{GPa}\) and 1.6 GPa, respectively (values given in the article "Influence of Material Properties Variability on Springback and Thinning in Sheet Stamping Processes: A Stochastic Analysis" (IntL. \(J\). of Advanced Manuf. Tech., 2010: 117-134)). a. If \(\bar{X}\) is the sample mean Young's modulus for a random sample of \(n=16\) sheets, where is the sampling distribution of \(\bar{X}\) centered, and what is the standard deviation of the \(\bar{X}\) distribution? b. Answer the questions posed in part (a) for a sample size of \(n=64\) sheets. c. For which of the two random samples, the one of part (a) or the one of part (b), is \(\bar{X}\) more likely to be within \(1 \mathrm{GPa}\) of \(70 \mathrm{GPa}\) ? Explain your reasoning.

Short Answer

Expert verified
The sampling distribution is centered at 70 GPa for both samples. The standard deviation is 0.4 GPa for n=16 and 0.2 GPa for n=64. Sample of n=64 is more likely to fall within 1 GPa of 70 GPa.

Step by step solution

01

Understand the Problem

We need to analyze the sampling distribution of the sample mean (\(\bar{X}\)) for different sample sizes. We'll find where the distribution is centered and determine its standard deviation for given sample sizes 16 and 64.
02

Define the Sampling Distribution Center

The center of the sampling distribution of \(\bar{X}\) is the population mean. Since the mean Young's modulus is given as \(70 \text{ GPa}\), the sampling distribution for any sample size is centered at \(70 \text{ GPa}\).
03

Calculate Standard Deviation for n=16

The standard deviation of the sampling distribution of \(\bar{X}\) when \(n=16\) is found using the formula:\[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\]Here, \(\sigma = 1.6 \text{ GPa}\) and \(n = 16\). Thus,\[\sigma_{\bar{X}} = \frac{1.6}{\sqrt{16}} = 0.4 \text{ GPa}\]
04

Calculate Standard Deviation for n=64

For \(n=64\), using the same formula, we find:\[\sigma_{\bar{X}} = \frac{1.6}{\sqrt{64}} = 0.2 \text{ GPa}\]
05

Compare Likelihood Within 1 GPa for Different Sample Sizes

With a smaller standard deviation, \(\bar{X}\) is more tightly centered around the mean. Hence, for \(n=64\), \(\bar{X}\) is more likely to be within \(1 \text{ GPa}\) of \(70 \text{ GPa}\) than for \(n=16\) because its distribution is narrower.

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.

Young's modulus
Young's modulus is a crucial concept in understanding the mechanical properties of materials. It quantifies the stiffness of an elastic material, essentially measuring how much a material will stretch or compress under a load. This is directly related to the stress and strain experienced by the material. If you imagine a spring, a higher Young's modulus means the spring is stiffer and doesn't stretch very much under the same amount of force. For aluminum alloy sheets, we are given a mean Young's modulus of 70 GPa, which tells us about the typical stiffness for this material type. A standard deviation of 1.6 GPa indicates variability in stiffness across different samples of this material. A material with a high Young's modulus is excellent for applications requiring minimal deformation under stress, such as in construction or manufacturing contexts.
Standard deviation
The standard deviation is a key statistic that measures the amount of variability or spread in a set of data points. A smaller standard deviation means that the data points tend to be closer to the mean, while a larger standard deviation indicates more spread out data. In our example with aluminum alloy sheets, the standard deviation is 1.6 GPa, showing how much the Young's modulus can vary between different sheets. Think of it like the range of possible stiffness values; the tighter this range, the more predictable the material's behavior. Calculating the standard deviation of the sample mean (often denoted as \( \sigma_{\bar{X}} \)) helps us understand how the average stiffness of a group of sheets might vary if we repeatedly took many samples. This involves dividing the population standard deviation by the square root of the sample size \( n \).
Sample mean
The sample mean \( \bar{X} \) represents the average value of a set of observations from a sample taken from a larger population. In the case of Young's modulus for aluminum sheets, the sample mean helps estimate the central tendency of stiffness for randomly selected sheets. For any given sample, the sample mean is influenced both by the true population mean and the variability intrinsic to the sample itself. The formula \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \) is used to calculate the standard deviation of the sample mean. This formula shows that as we increase the sample size \( n \), the standard deviation of the sample mean decreases, indicating that our estimates of the population mean become more accurate. This is why for a larger sample size like \( n=64 \), the sample mean gives us a more precise measure than for \( n=16 \).
Central limit theorem
The central limit theorem (CLT) is a fundamental principle in statistics. It states that the distribution of the sample mean will approximate a normal distribution, regardless of the shape of the population distribution, provided the sample size is sufficiently large. This theorem is particularly powerful because it allows statisticians to make inferences about population parameters from sample statistics. In simpler terms, even if the data from individual aluminum sheets isn't normally distributed, the averages (means) of these samples will form a normal distribution as the number of samples increases. The CLT provides the foundation for understanding why, with larger samples like \( n=64 \), the distribution of the sample mean is more likely to be closer to the true population mean, making our predictions and estimations much more reliable.

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

The National Health Statistics Reports dated Oct. 22, 2008, stated that for a sample size of 277 18-year-old American males, the sample mean waist circumference was \(86.3 \mathrm{~cm}\). A somewhat complicated method was used to estimate various population percentiles, resulting in the following values: \(\begin{array}{lllllll}5^{\text {th }} & 10^{\text {th }} & 25^{\text {th }} & 50^{\text {th }} & 75^{\text {th }} & 90^{\text {th }} & 95^{\text {th }} \\\ 69.6 & 70.9 & 75.2 & 81.3 & 95.4 & 107.1 & 116.4\end{array}\) a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution. b. Suppose that the population mean waist size is \(85 \mathrm{~cm}\) and that the population standard deviation is \(15 \mathrm{~cm}\). How likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least \(86.3 \mathrm{~cm}\) ? c. Referring back to (b), suppose now that the population mean waist size in \(82 \mathrm{~cm}\). Now what is the (approximate) probability that the sample mean will be at least \(86.3 \mathrm{~cm}\) ? In light of this calculation, do you think that \(82 \mathrm{~cm}\) is a reasonable value

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a certain region is normally distributed with mean \(2.65\) and standard deviation .85 (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants," Water Research, 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least 99 ?

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3)\), (2) \(125 \mathrm{ft}^{3}\), and (3) \(512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) denote the number of type \(i\) containers shipped during a given week. With \(\mu_{i}=E\left(X_{i}\right)\) and \(\sigma_{i}^{2}=V\left(X_{i}\right)\), suppose that the mean values and standard deviations are as follows: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume \(=27 X_{1}+125 X_{2}+512 X_{3}\).] b. Would your calculations necessarily be correct if the \(X_{i}^{\prime}\) s were not independent? Explain.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be random variables denoting \(n\) independent bids for an item that is for sale. Suppose each \(X_{i}\) is uniformly distributed on the interval \([100,200]\). If the seller sells to the highest bidder, how much can he expect to earn on the sale? [Hint: Let \(Y=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right)\). First find \(F_{r}(y)\) by noting that \(Y \leq y\) iff each \(X_{i}\) is \(\leq y\). Then obtain the pdf and \(E(Y) .]\)

Let \(X\) denote the courtship time for a randomly selected female-male pair of mating scorpion flies (time from the beginning of interaction until mating). Suppose the mean value of \(X\) is \(120 \mathrm{~min}\) and the standard deviation of \(X\) is \(110 \mathrm{~min}\) (suggested by data in the article "Should I Stay or Should I Go? Condition- and Status-Dependent Courtship Decisions in the Scorpion Fly Panorpa Cognate" (Animal Behavior, 2009: 491-497)). a. Is it plausible that \(X\) is normally distributed? b. For a random sample of 50 such pairs, what is the (approximate) probability that the sample mean courtship time is between 100 min and 125 min? c. For a random sample of 50 such pairs, what is the (approximate) probability that the total courtship time exceeds 150 hr? d. Could the probability requested in (b) be calculated from the given information if the sample size were 15 rather than 50 ? Explain.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Probability and 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.