/*! 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 The breaking strength of a rivet... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 50

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200 ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?

Short Answer

Expert verified
a) The probability is approximately 0.8905. b) No, the sample size of 15 is too small to assume normality.

Step by step solution

01

Identify the Problem

We are given the mean breaking strength of a rivet as \( \mu = 10,000 \) psi, standard deviation \( \sigma = 500 \) psi, and a sample size of 40. We need to find the probability that the sample mean is between 9900 and 10,200 psi.
02

Calculate Standard Error

To find the probability, we first need the standard error (SE) of the sample mean, which is calculated by \( SE = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the sample size.
03

Compute Standard Error with n=40

Substitute \( n = 40 \) and \( \sigma = 500 \) into the SE formula: \[ SE = \frac{500}{\sqrt{40}} \approx 79.06 \] psi.
04

Standardize the Boundary Values

To find the probability, convert the boundary values 9900 and 10,200 to standard normal variables (z-scores) using \[ z = \frac{\bar{x} - \mu}{SE} \]. This converts the sample mean limits into z-scores.
05

Calculate Z-scores for 9900 and 10,200

For 9900: \[ z = \frac{9900 - 10000}{79.06} \approx -1.26 \] For 10,200: \[ z = \frac{10200 - 10000}{79.06} \approx 2.53 \]
06

Find Probability using Z-scores

Use a standard normal distribution table (or a calculator) to find the probability corresponding to the z-scores. The probability that the mean is between the z-scores is given by: \[ P(-1.26 < Z < 2.53) = P(Z < 2.53) - P(Z < -1.26) \approx 0.9943 - 0.1038 = 0.8905 \]
07

Consider Sample Size Change to 15

For part (b), if the sample size is 15, the finite sample size is too small for the Central Limit Theorem to assure normality. Thus, with n = 15, we cannot use this method to calculate the probability.

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.

Standard Error
Standard Error (SE) is a crucial concept in statistics. It provides a measure of how much the sample mean is expected to vary from the true population mean. Think of it as the "margin of error" for your sample mean. To calculate SE, you use the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \]where:
  • \(\sigma\) is the standard deviation of the population.
  • \(n\) is the sample size.
In the example exercise, we found the SE for a sample size of 40 with a population standard deviation of 500 psi. The SE turned out to be approximately 79.06 psi. This means the sample mean breaking strength may deviate by about 79.06 psi from the actual mean of 10,000 psi.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It's given in terms of standard deviations from the mean. To find a Z-score, you use the following formula: \[ z = \frac{\bar{x} - \mu}{SE} \]where:
  • \(\bar{x}\) is the sample mean you are interested in.
  • \(\mu\) is the population mean.
  • \(SE\) is the standard error.
In our example, we calculated the Z-scores for boundary values 9900 and 10200 psi. The calculated Z-scores were -1.26 and 2.53, respectively. This standardization allows you to determine the probability from a standard normal distribution table. It shows how extreme or typical the sample mean is within the context of the population.
Central Limit Theorem
The Central Limit Theorem (CLT) is a cornerstone of statistics. It states that if you have a large enough sample size, the distribution of the sample mean will be approximately normal (or bell-shaped), regardless of the shape of the population distribution. This is especially helpful because it allows us to make inferences about the population mean using the properties of a normal distribution. In practical terms, this means:
  • For a sample size as small as 30, the distribution of sample means will often be approximately normal.
  • With smaller samples, like the size 15 in part b of the exercise, this normal approximation might not be reliable.
Because of the CLT, we could use normal probability calculations for our sample size of 40. However, it wouldn鈥檛 work as well with only 15 samples, making calculations less reliable.
Sample Size
Sample size, denoted as \(n\), is the number of observations in your sample. It's one of the most important factors affecting statistical reliability. Bigger sample sizes lead to more precise estimates of the population mean, narrowing down the margin of standard error.Here's why sample size matters:
  • A larger \(n\) generally results in a smaller standard error, leading to more accurate and stable sample means.
  • With the Central Limit Theorem, as \(n\) increases, the distribution of the sample mean becomes more normal鈥攅nabling more valid normal approximation.
In this exercise, a sample size of 40 allowed us to use normal distribution probabilities to find the sample mean's likelihood. With a smaller sample of 15, this approach wasn鈥檛 feasible due to increased variability and a non-normally distributed mean.

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

A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1,10 by supplier 2 , and 12 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let \(X=\) the number of supplier l's components selected, \(Y=\) the number of supplier 2 's components selected, and \(p(x, y)\) denote the joint pmf of \(X\) and \(Y\). a. What is \(p(3,2)\) ? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, \(p(3,2)=\) (number of outcomes with \(X=3\) and \(Y=2\) )/(total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] b. Using the logic of part (a), obtain \(p(x, y)\). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

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 ?

The joint probability distribution of the number \(X\) of cars and the number \(Y\) of buses per signal cycle at a proposed left-turn lane is displayed in the accompanying joint probability table. \begin{tabular}{cc|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline & 0 & \(.025\) & \(.015\) & \(.010\) \\ & 1 & \(.050\) & \(.030\) & \(.020\) \\ & 2 & \(.125\) & \(.075\) & \(.050\) \\ \(x\) & 3 & \(.150\) & \(.090\) & \(.060\) \\ & 4 & \(.100\) & \(.060\) & \(.040\) \\ & 5 & \(.050\) & \(.030\) & \(.020\) \end{tabular} a. What is the probability that there is exactly one car and exactly one bus during a cycle? b. What is the probability that there is at most one car and at most one bus during a cycle? c. What is the probability that there is exactly one car during a cycle? Exactly one bus? d. Suppose the left-turn lane is to have a capacity of five cars, and that one bus is equivalent to three cars. What is the probability of an overflow during a cycle? e. Are \(X\) and \(Y\) independent rv's? Explain.

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5 \mathrm{~min}\), respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

Let \(X_{1}, X_{2}\) and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=15\), calculate \(P\left(T_{o} \leq 200\right)\) and \(P\left(150 \leq T_{a} \leq 200\right) ?\) b. Using the \(\mu_{i}\) 's and \(\sigma_{i}^{\prime}\) s given in part (a), calculate both \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate and interpret \(P\left(-10 \leq X_{1}-.5 X_{2}-5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \mu_{2}=50, \mu_{3}=60, \sigma_{1}^{2}=10, \sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 160\right)\) and also \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right) .\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

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.