/*! 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 17 A manufacturing process is desig... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturing process is designed to produce bolts with a diameter of 0.5 inches. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation of bolt diameters is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the actual process mean is 0.5 inches.)

Short Answer

Expert verified
To find the probability of unnecessary shutdown, first calculate the Z-scores for \(\bar{x}_{min} = 0.49\) inches and \(\bar{x}_{max} = 0.51\) inches using the formula: \[Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]. After calculating the Z-scores, use the standard normal distribution table to find the probabilities for each Z-score. Finally, add these two probabilities to get the total probability of unnecessary shutdown. \[P(\text{shutdown unnecessarily}) = P(Z < Z_{min}) + P(Z > Z_{max})\]

Step by step solution

01

Identify the given values

We are given the following information: - Sample size: \(n = 36\) - Process Mean: \(\mu = 0.5\) inches - Sample Mean Range for shutdown: \(\bar{x}_{min} = 0.49\) inches and \(\bar{x}_{max} = 0.51\) inches - Standard Deviation: \(\sigma = 0.02\) inches
02

Calculate the Z-scores

Next, we'll need to find the Z-score for both \(\bar{x}_{min}\) and \(\bar{x}_{max}\), which is a measure of how many standard deviations away from the mean a certain value is. We use the formula: \[Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\] For \(\bar{x}_{min}\): \[Z_{min} = \frac{0.49 - 0.5}{\frac{0.02}{\sqrt{36}}}\] For \(\bar{x}_{max}\): \[Z_{max} = \frac{0.51 - 0.5}{\frac{0.02}{\sqrt{36}}}\]
03

Calculate the probabilities

Now we are going to use the Z-scores we calculated to find the probability of the manufacturing line being shut down due to a falsely low or high sample mean. To do this, we'll refer to the standard normal distribution table (Z-table). First, find the probability of shutting down due to falsely low sample mean by using \(Z_{min}\): \[P(Z < Z_{min})\] Next, find the probability of shutting down due to falsely high sample mean by using \(Z_{max}\): \[P(Z > Z_{max})\]
04

Find the probability of shutting down unnecessarily

Now that we have the probabilities for shutting down due to falsely low and high sample means, we can find the total probability of shutting down unnecessarily. \[P(\text{shutdown unnecessarily}) = P(Z < Z_{min}) + P(Z > Z_{max})\] This will give us the probability of the manufacturing line being shut down unnecessarily.

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 91Ó°ÊÓ!

Key Concepts

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

Z-score
The Z-score is a statistical measure that tells us how far a specific data point is from the mean of the data set, measured in terms of standard deviations. In other words, it helps us understand the position of a sample mean relative to the process mean. Using the formula \( Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \), we can calculate the Z-score for a sample mean, where:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the process mean
  • \(\sigma\) is the standard deviation of the population
  • \(n\) is the sample size
This process lets us determine the number of standard deviations a sample mean is from the process mean, which is crucial for determining the likelihood of observing such a sample mean under normal process conditions. Understanding Z-scores thus helps in decision-making processes like determining whether a manufacturing procedure needs adjustment.
Standard Deviation
Standard deviation (\(\sigma\)) is a measure that quantifies the amount of variation or dispersion in a set of data values. A low standard deviation indicates that the data points are close to the mean, while a high standard deviation indicates a greater spread of data points. In the context of the manufacturing process exercise, the standard deviation is crucial as it helps define the expected variability in bolt diameters.When calculating Z-scores for the sample mean, the standard deviation plays a key role in the formula. It helps us standardize the difference between the sample mean and the process mean to decide how unusual or normal a sample observation is.A good understanding of standard deviation allows us to interpret variability in a process effectively, which in turn aids in controlling and improving quality in production settings.
Sample Mean
The sample mean (\(\bar{x}\)) represents the average value from a sample taken from a population. It's a critical statistic because it provides a snapshot of the larger group’s characteristics and is used to estimate the population mean. In the exercise, the sample mean of 36 bolts is used to check if the process mean deviates enough to warrant shutting down the manufacturing line.Monitoring sample means is essential in quality control processes because it allows for real-time adjustments to maintain standards. If the sample mean falls outside of acceptable limits (such as less than 0.49 inches or more than 0.51 inches in our exercise), it indicates a potential problem that might need immediate correction.Understanding the concept of sample mean helps in assessing group performance and guiding decisions based on sampled data.
Probability
Probability in statistics measures the likelihood of a particular event occurring. In our exercise, it's used to calculate the chance that the sample mean falls outside the acceptable range, prompting a shutdown of the production line. To calculate this probability, we use the standard normal distribution, which involves utilizing Z-scores to find the corresponding probability values. Determining these probabilities involves:
  • Calculating the likelihood that the sample mean is below the minimum threshold (< 0.49 inches)
  • Calculating the likelihood that it is above the maximum threshold (> 0.51 inches)
  • Combining these probabilities to understand the overall risk of unnecessary shutdowns
Mastering probability helps in making informed decisions by quantifying uncertainty, an essential aspect of managing and improving a process through statistical control methods.

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

Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination ("Study Goals and Procrastination Tendencies at Different Stages of the Undergraduate Degree," Studies in Higher Education [2016]: 2028-2043). Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to \(64,\) with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 37.02 and the standard deviation was 6.44 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean procrastination scale for first-year students at this college. (Hint: See Example 12.7.) b. Based on your interval, is 40 a plausible value for the population mean score? What does this imply about the population of first-year students?

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{ll}30.3 & 29.6\end{array}\) \(\begin{array}{llll}27.2 & 29.3 & 31.2 & 28.4\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? Describe the shape of the sampling distribution of \(\bar{x}\). b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

A random sample is selected from a population with mean \(\mu=60\) and standard deviation \(\sigma=3\). Determine the mean and standard deviation of the sampling distribution of \(\bar{x}\) for each of the following sample sizes: a. \(n=6\) d. \(n=75\) b. \(n=18\) e. \(n=200\) c. \(n=42\) f. \(n=400\)

See all solutions

Recommended explanations on Math Textbooks

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.