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Chapter 4: Problem 18

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is 500 in. A bearing is acceptable if its diameter is within .004 in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with mean value \(.499\) in. and standard deviation .002 in. What percentage of the bearings produced will not be acceptable?

Short Answer

Expert verified
7.3% of the bearings will not be acceptable.

Step by step solution

01

Define the Acceptable Range

The bearings are acceptable if their diameters are within \(.004\) inches of the target diameter of \(0.500\) inches. Therefore, the acceptable diameter range is from \(0.500 - 0.004 = 0.496\) inches to \(0.500 + 0.004 = 0.504\) inches.
02

Use the Z-score Formula

To find the percentage of bearings not within the acceptable range, we need to calculate the Z-scores for the boundary diameters. The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the boundary diameter, \( \mu = 0.499 \) is the mean, and \( \sigma = 0.002 \) is the standard deviation.
03

Compute Z-scores for the Lower Bound

For the lower bound of the acceptable range, \( X = 0.496 \) inches: \[ Z = \frac{0.496 - 0.499}{0.002} = \frac{-0.003}{0.002} = -1.5 \]
04

Compute Z-scores for the Upper Bound

For the upper bound of the acceptable range, \( X = 0.504 \) inches: \[ Z = \frac{0.504 - 0.499}{0.002} = \frac{0.005}{0.002} = 2.5 \]
05

Find the Probability Using the Standard Normal Distribution Table

The Z-score of \(-1.5\) correlates with a probability of approximately \(0.0668\), and for a Z-score of \(2.5\), the probability is approximately \(0.9938\). This tells us the probabilities of being below \(0.496\) and below \(0.504\).
06

Calculate the Probability of Being Within the Range

The probability that a bearing's diameter is within the acceptable range \([0.496, 0.504]\) is the difference between the two probabilities: \[ 0.9938 - 0.0668 = 0.927 \]
07

Determine the Probability of an Unacceptable Bearing

The total probability of a bearing not being within the acceptable range is the complement of the calculated in-range probability: \[ 1 - 0.927 = 0.073 \] or \(7.3\%\).

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Key Concepts

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

Z-score
A Z-score is a statistical measurement that tells us how far a particular value is from the mean of a distribution in terms of standard deviations. It's a way of standardizing scores on the same scale, allowing for a clear comparison.
  • The formula for calculating the Z-score is given by: \( Z = \frac{X - \mu}{\sigma} \).
  • Here, \( X \) represents the specific value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the context of the exercise, Z-scores help determine how far the specific diameter of a bearing is from the mean diameter \(0.499\) inches. Using the calculation, the Z-scores for the lower and upper boundaries of acceptable diameters can be found. Obtaining these scores is essential for probability calculations in the next steps.
Standard Deviation
Standard deviation is a critical concept in statistics that measures the amount of variation or dispersion in a set of values. A low standard deviation implies that the values tend to be close to the mean of the data set, whereas a high standard deviation implies a broad range of values.
  • In our exercise, the standard deviation is \(0.002\) inches.
  • This small value indicates that most ball bearings have diameters close to the mean \(0.499\) inches.
Since bearings with diameters just slightly deviating from the mean (more than \(0.004\) inches) are considered unacceptable, having a precise standard deviation allows us to predict the likelihood of such deviations.
Probability Calculation
Probability calculations are used to determine the likelihood of a particular event. In this scenario, we use the standard normal distribution's properties to find the probabilities associated with different Z-scores.
  • We start by calculating the Z-scores for both the lower (\(0.496\) inches) and upper (\(0.504\) inches) bounds of the acceptable range.
  • Then, we find the probability of scoring below each bound using standard normal distribution tables.
  • For the Z-scores of -1.5 and 2.5, the probabilities are approximately \(0.0668\) and \(0.9938\) respectively.
By computing the difference between these probabilities, we obtain the probability that a bearing's diameter is within the acceptable range. This calculation helps determine the percentage of bearings that meet the desired specifications.
Boundary Conditions
Boundary conditions in this context define the acceptable range for the diameter of ball bearings. Establishing these limits is crucial to ensure quality control.
  • The target diameter is \(0.500\) inches, which helps set the acceptable range from \(0.496\) to \(0.504\) inches.
  • The change in machine settings leads to a shift in the mean diameter to \(0.499\) inches, which affects the percentage of acceptable bearings.
Understanding these boundary conditions allows us to set the parameters for when a bearing is rejected. It's these conditions that determine the overall quality and performance standards in the production process, effectively helping in maintaining consistency and minimizing defects.

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Most popular questions from this chapter

Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

The article "The Statistics of Phytotoxic Air Pollutants" (J. of Royal Stat. Soc., 1989: 183-198) suggests the lognormal distribution as a model for \(\mathrm{SO}_{2}\) concentration above a certain forest. Suppose the parameter values are \(\mu=1.9\) and \(\sigma=.9\). a. What are the mean value and standard deviation of concentration? b. What is the probability that concentration is at most 10 ? Between 5 and 10 ?

Let \(X\) have the Pareto pdf $$ f(x ; k, \theta)=\left\\{\begin{array}{cl} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ introduced in Exercise \(10 .\) a. If \(k>1\), compute \(E(X)\). b. What can you say about \(E(X)\) if \(k=1\) ? c. If \(k>2\), show that \(V(X)=k \theta^{2}(k-1)^{-2}(k-2)^{-1}\). d. If \(k=2\), what can you say about \(V(X)\) ? e. What conditions on \(k\) are necessary to ensure that \(E\left(X^{n}\right)\) is finite?

Nonpoint source loads are chemical masses that travel to the main stem of a river and its tributaries in flows that are distributed over relatively long stream reaches, in contrast to those that enter at well-defined and regulated points. The article "Assessing Uncertainty in Mass Balance Calculation of River Nonpoint Source Loads" (J. of Envir. Engr., 2008: 247-258) suggested that for a certain time period and location, \(X\) = nonpoint source load of total dissolved solids could be modeled with a lognormal distribution having mean value \(10,281 \mathrm{~kg} / \mathrm{day} / \mathrm{km}\) and a coefficient of variation \(C V=.40\left(C V=\sigma_{X} / \mu_{X}\right)\). a. What are the mean value and standard deviation of \(\ln (X) ?\) b. What is the probability that \(X\) is at most 15,000 \(\mathrm{kg} / \mathrm{day} / \mathrm{km}\) ? c. What is the probability that \(X\) exceeds its mean value, and why is this probability not \(.5\) ? d. Is 17,000 the 95 th percentile of the distribution?

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, 1984: 1169-1174) suggests the uniform distribution on the interval \((7.5,20)\) as a model for depth (cm) of the bioturbation layer in sediment in a certain region. a. What are the mean and variance of depth? b. What is the cdf of depth? c. What is the probability that observed depth is at most 10 ? Between 10 and 15 ? d. What is the probability that the observed depth is within 1 standard deviation of the mean value? Within 2 standard deviations?
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