/*! 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 50 Shear strength measurements for ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 50

Shear strength measurements for spot welds have been found to have standard deviation 10 pounds per square inch (psi). If 100 test welds are to be measured, what is the approximate probability that the sample mean will be within 1 psi of the true population mean?

Short Answer

Expert verified
The probability is approximately 0.6826.

Step by step solution

01

Identify Given Information

We are given that the standard deviation \( \sigma \) of the shear strength is 10 psi, and we have a sample size \( n = 100 \). We need to find the probability that the sample mean \( \bar{x} \) is within 1 psi of the population mean \( \mu \).
02

Calculate the Standard Error

The standard error (SE) of the sample mean is calculated using the formula \( \text{SE} = \frac{\sigma}{\sqrt{n}} \). Plugging in the values, we get \( \text{SE} = \frac{10}{\sqrt{100}} = 1 \) psi.
03

Use the Central Limit Theorem

The Central Limit Theorem states that the sampling distribution of the sample mean is approximately normal if the sample size is large. Here, the sample size is 100, so we can assume a normal distribution with mean \( \mu \) and standard deviation \( \text{SE} = 1 \) psi.
04

Define the Range of Interest

We are asked for the probability that the sample mean is within 1 psi of the true mean. This means we need \( \mu - 1 \leq \bar{x} \leq \mu + 1 \).
05

Calculate the Cumulative Probability

Since \( \bar{x} \) follows a normal distribution with mean \( \mu \) and standard deviation \( \text{SE} = 1 \), we calculate the probability \( P(\mu - 1 \leq \bar{x} \leq \mu + 1) \). This is equivalent to finding \( P(-1 \leq Z \leq 1) \) where \( Z \) is a standard normal variable. Using standard normal distribution tables or a calculator, we know \( P(-1 \leq Z \leq 1) = 0.6826 \).
06

Finalize the Result

Therefore, the approximate probability that the sample mean is within 1 psi of the true population mean is 0.6826.

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.

Shear Strength Measurements
Shear strength measurements are crucial in evaluating the integrity of spot welds, which are widely used in various industrial applications. Spot welds are essential in joining metal pieces, and their strength is typically measured in pounds per square inch (psi).
This measurement tells us how much force a weld can withstand before failing. Understanding shear strength is important because it affects the durability and safety of welded structures.
  • High shear strength signifies a strong weld capable of withstanding rigorous use.
  • Low shear strength might indicate potential issues, such as defects or inadequate welding processes.
In our context, the shear strength measurements have a known standard deviation of 10 psi. This allows us to quantify the variability and reliability of the welding process across different samples.
Standard Error
The standard error (SE) is a statistical term that measures the accuracy with which a sample represents a population. It is especially useful when dealing with sample means. The SE tells us how much variation can be expected when estimating the population mean from a sample.
In our exercise, the standard error is calculated using the formula:\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]Where:
  • \( \sigma \) is the population standard deviation (10 psi in our case).
  • \( n \) is the sample size (100 test welds).
When we plug in our values, we find the SE to be 1 psi. This indicates that the sample mean should typically be within 1 psi of the true population mean, providing a concrete measure of sampling variation.
Normal Distribution
The normal distribution is a fundamental concept in statistics, often referred to as a 'bell curve'. It is characterized by its symmetrical shape, where most observations cluster around the mean. In the context of sampling, understanding normal distribution is key to applying the Central Limit Theorem.
According to the Central Limit Theorem, the sampling distribution of the sample mean approximates a normal distribution, especially when our sample size is large, like the 100 test welds here. This gives us a powerful way to make inferences about the population mean:
  • The mean of the sampling distribution is the same as the population mean (\( \mu \)).
  • The standard deviation of the sampling distribution is the standard error (\( \text{SE} \)), calculated previously as 1 psi.
This allows us to determine the probability of the sample mean falling within a specific range.
Sample Mean Probability
Determining the sample mean's probability involves establishing how likely the sample mean is to fall within a certain range of the population mean. This is crucial when assessing the reliability of sample data.
For our problem, we're interested in finding the probability that the sample mean is within 1 psi of the population mean. This can be visualized using the standard normal distribution (Z-distribution).
  • We look for the probability between \( -1 \) and \( 1 \) in a standard normal distribution, since these correspond to being within 1 SE of the mean.
Statistical tools or tables reveal that \( P(-1 \leq Z \leq 1) = 0.6826 \). Therefore, there is a 68.26% chance that our sample mean will lie within 1 psi of the true mean, giving us confidence in our measurements.

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

Suppose that \(Z\) has a standard normal distribution and that \(Y\) is an independent \(\chi^{2}\) -distributed random variable with \(\nu\) df. Then, according to Definition 7.2, $$T=\frac{Z}{\sqrt{Y / \nu}}$$ has a \(t\) distribution with \(\nu\) df. \(^{*}\) a. If \(Z\) has a standard normal distribution, give \(E(Z)\) and \(E\left(Z^{2}\right)\). [Hint: For any random variable, \(\left.E\left(Z^{2}\right)=V(Z)+(E(Z))^{2} \cdot\right]\) b. According to the result derived in Exercise \(4.112(\mathrm{a}),\) if \(Y\) has a \(\chi^{2}\) distribution with \(\nu\) df, then $$E\left(Y^{a}\right)=\frac{\Gamma([\nu / 2]+a)}{\Gamma(\nu / 2)} 2^{a}, \quad \text { if } \nu>-2 a$$. Use this result, the result from part (a), and the structure of \(T\) to show the following. [Hint: Recall the independence of \(Z \text { and } Y \)] i. \(E(T)=0,\) if \(\nu>1\) ii. \(V(T)=\nu /(\nu-2),\) if \(\nu>2\)

The efficiency (in lumens per watt) of light bulbs of a certain type has population mean 9.5 and standard deviation. \(5,\) according to production specifications. The specifications for a room in which eight of these bulbs are to be installed call for the average efficiency of the eight bulbs to exceed 10. Find the probability that this specification for the room will be met, assuming that efficiency measurements are normally distributed.

In 2004 Florida was hit by four major hurricanes. In 2005 a survey indicated that, in \(2004,48 \%\) of the households in Florida had no plans for escaping an approaching hurricane. Suppose that a recent random sample of 50 households was selected in Gainesville and that those in 29 of the households indicated that their household had a hurricane escape plan. a. If the 2004 state percentages still apply to recent Gainesville households, use the Normal Approximation to Binomial Distribution applet to find the exact and approximate values of the probability that 29 or more of the households sampled have a hurricane escape plan. b. Refer to part (a). Is the normal approximation close to the exact binomial probability? Explain why.

Just as the difference between two sample means is normally distributed for large samples, so is the difference between two sample proportions. That is, if \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables with parameters \(\left(n_{1}, p_{1}\right)\) and \(\left(n_{2}, p_{2}\right),\) respectively, then \(\left(Y_{1} / n_{1}\right)-\left(Y_{2} / n_{2}\right)\) is approximately normally distributed for large values of \(n_{1}\) and \(n_{2}\) a. Find \(E\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\) b. Find \(V\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\)

An experimenter is comparing two methods for removing bacteria colonies from processed luncheon meats. After treating some samples by method A and other identical samples by method B, the experimenter selects a 2 -cubic-centimeter subsample from each sample and makes bacteria colony counts on these subsamples. Let \(X\) denote the total count for the subsamples treated by method A and let \(Y\) denote the total count for the subsamples treated by method B. Assume that \(X\) and \(Y\) are independent Poisson random variables with means \(\lambda_{1}\) and \(\lambda_{2}\), respectively. If \(X\) exceeds Y by more than \(10,\) method \(\mathrm{B}\) will be judged superior to A. Suppose that, in fact, \(\lambda_{1}=\lambda_{2}=50 .\) Find the approximate probability that method B will be judged superior to method A.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Applied Mathematics

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.