/*! 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 14 A manufacturer of bolts has a qu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturer of bolts has a quality control policy that requires it to destroy any bolts that are more than 2 standard deviations from the mean. The quality-control engineer knows that the bolts coming off the assembly line have a mean length of \(8 \mathrm{~cm}\) with a standard deviation of \(0.05 \mathrm{~cm} .\) For what lengths will a bolt be destroyed?

Short Answer

Expert verified
Bolts shorter than 7.9 cm or longer than 8.1 cm.

Step by step solution

01

Understand the Given Information

Identify the given quantities: the mean length of the bolts is \(8 \mathrm{~cm}\) and the standard deviation is \(0.05 \mathrm{~cm}\). The quality control policy states that bolts more than 2 standard deviations from the mean will be destroyed.
02

Calculate the Range for Acceptable Lengths

Calculate the allowable range by adding and subtracting 2 standard deviations from the mean: 1. Lower limit: \(\mu - 2\sigma = 8 - 2(0.05)\)2. Upper limit: \(\mu + 2\sigma = 8 + 2(0.05)\).
03

Perform the Calculations

1. Lower limit: \( 8 - 2(0.05) = 8 - 0.1 = 7.9 \mathrm{~cm} \)2. Upper limit: \( 8 + 2(0.05) = 8 + 0.1 = 8.1 \mathrm{~cm} \)
04

Determine the Destruction Range

Bolts that are shorter than 7.9 cm or longer than 8.1 cm will be destroyed.

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.

standard deviation
Standard deviation measures how spread out the numbers in a data set are. It tells us how much variation or dispersion there is from the average (mean). For example, in manufacturing, if you produce bolts that should be 8 cm on average, the standard deviation helps you understand how much individual bolt lengths vary from this average.
Mathematically, standard deviation is represented as \[ \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2} \] where \( \sigma \) is the standard deviation, \( N \) is the number of data points, \( x_i \) represents each data point, and \( \mu \) is the mean. The larger the standard deviation, the more spread out the data points are.
In our exercise, the standard deviation of the bolt lengths is 0.05 cm, indicating minor variation from the mean length of 8 cm.
mean calculation
Mean calculation is a fundamental concept in statistics. The mean is the average of a set of numbers. It’s calculated by adding all the values together and then dividing by the number of values. For the bolts, if you want to know the average length, you sum up the lengths of all bolts and divide by the number of bolts.
Mathematically, this is expressed as \[ \mu = \frac{1}{N}\sum_{i=1}^{N}x_i \] where \( \mu \) is the mean, \( N \) is the number of data points, and \( x_i \) represents each data point. In our example, the mean length of the bolts is given as 8 cm. This mean value helps to set a standard for comparison with individual bolt lengths.
acceptable range determination
Determining the acceptable range is crucial in quality control to decide which products to keep. Here's how you do it: First, you need the mean and standard deviation of your data, which are 8 cm and 0.05 cm for our bolts. The quality control policy specifies that bolts differing by more than 2 standard deviations from the mean will be destroyed.
To calculate the acceptable range:
  • First, find the total deviation allowed by multiplying the standard deviation (0.05 cm) by 2: \(2 \times 0.05 = 0.1\) cm.
  • Next, determine the lower limit by subtracting the total deviation from the mean: \(8 - 0.1 = 7.9\) cm.
  • Then, find the upper limit by adding the total deviation to the mean: \(8 + 0.1 = 8.1\) cm.
Therefore, any bolt shorter than 7.9 cm or longer than 8.1 cm will be destroyed.
statistical quality control
Statistical quality control (SQC) involves using statistical methods to monitor and control production processes. It ensures products are manufactured to specific standards consistently. A major part of SQC is setting acceptable ranges and identifying products that do not meet these standards.
In our example, the manufacturer uses the mean and standard deviation to set tolerance levels. By detecting whether bolts fall within 2 standard deviations of the mean, the manufacturer can ensure most bolts are within the desired length.
SQC tools like control charts, process capability analysis, and histograms can help in analyzing data and sustaining quality. Employing SQC helps to reduce waste, minimize defects, and maintain product reliability and customer satisfaction.

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

True or False: When comparing two populations, the larger the standard deviation, the more dispersion the distribution has. provided that the variable of interest from the two populations has the same unit of measure.

Violent crimes include rape, robbery, assault, and homicide. The following is a summary of the violent-crime rate (violent crimes per 100,000 population) for all 50 states in the United States plus Washington, D.C., in 2012 . $$ Q_{1}=252.4 \quad Q_{2}=333.8 \quad Q_{3}=454.5 $$ (a) Provide an interpretation of these results. (b) Determine and interpret the interquartile range. (c) The violent-crime rate in Washington, D.C., in 2012 was 1243.7. Would this be an outlier? (d) Do you believe that the distribution of violent-crime rates is skewed or symmetric? Why?

The acidity or alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The following data represent the \(\mathrm{pH}\) in samples of bottled water and tap water. $$ \begin{array}{lllllll} \hline \text { Tap } & 7.64 & 7.45 & 7.47 & 7.50 & 7.68 & 7.69 \\ & 7.45 & 7.10 & 7.56 & 7.47 & 7.52 & 7.47 \\ \hline \text { Bottled } & 5.15 & 5.09 & 5.26 & 5.20 & 5.02 & 5.23 \\ & 5.28 & 5.26 & 5.13 & 5.26 & 5.21 & 5.24 \\ \hline \end{array} $$ (a) Determine the mean, median, and mode \(\mathrm{pH}\) for each type of water. Comment on the differences between the two water types. (b) Suppose the \(\mathrm{pH}\) of 7.10 in tap water was incorrectly recorded as \(1.70 .\) How does this affect the mean? the median? What property of the median does this illustrate?

Lawrence Summers (former Secretary of the Treasury and former president of Harvard) infamously claimed that women have a lower standard deviation IQ than men. He went on to suggest that this was a potential explanation as to why there are fewer women in top math and science positions. Suppose an IQ of 145 or higher is required to be a researcher at a top-notch research institution. Use the idea of standard deviation, the Empirical Rule, and the fact that the mean and standard deviation IQ of humans is 100 and 15 , respectively, to explain Summers' argument.

The data below represent the age of the mother at the time of her first birth for a random sample of 30 mothers. $$ \begin{array}{llllll} \hline 21 & 35 & 33 & 25 & 22 & 26 \\ \hline 21 & 24 & 16 & 32 & 25 & 20 \\ \hline 30 & 20 & 20 & 29 & 21 & 19 \\ \hline 18 & 24 & 33 & 22 & 23 & 25 \\ \hline 17 & 23 & 25 & 29 & 25 & 19 \\ \hline \end{array} $$ (a) Construct a box plot of the data. (b) Use the box plot and quartiles to describe the shape of the distribution.

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.