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A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(500 \mathrm{in}\). 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 produced will not be acceptable.

Step by step solution

01

Define the Acceptable Range

The acceptable diameter range for the bearings is from \(0.500 - 0.004 = 0.496\) in. to \(0.500 + 0.004 = 0.504\) in.
02

Set the Given Mean and Standard Deviation

The bearings now have a mean diameter of \(0.499\) in. and a standard deviation of \(0.002\) in.
03

Calculate the Z-scores for the Acceptable Range

The **Z-score** for a diameter of \(0.496\) is calculated as: \[Z = \frac{0.496 - 0.499}{0.002} = -1.5\]. The Z-score for a diameter of \(0.504\) is: \[Z = \frac{0.504 - 0.499}{0.002} = 2.5\].
04

Find the Probability of the Acceptable Range

Using the Z-table, the probability corresponding to \(Z = -1.5\) is approximately 0.0668, and for \(Z = 2.5\) it is approximately 0.9938. The probability of a bearing being acceptable is the difference: \(0.9938 - 0.0668 = 0.927\).
05

Determine the Percentage of Unacceptable Bearings

The percentage of bearings that are not acceptable is the complement of the probability of being acceptable. This is: \(1 - 0.927 = 0.073\), or 7.3%.

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Key Concepts

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

Understanding Z-score Calculation
The Z-score is an essential concept in statistics, especially when working with normally distributed data. It helps determine how far a particular data point is from the mean, measured in units of standard deviation. By understanding the Z-score, we gain insights into the probability of a data point occurring within a given set of normal distribution parameters.

The formula for calculating the Z-score is:
  • \[ Z = \frac{(X - \mu)}{\sigma} \]
where:
  • \(X\) is the value for which you want to find the Z-score.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
In our example, the Z-score helps assess how unusual or common a bearing's diameter is, relative to others produced by the machine. For diameters on the edge of the acceptable range \(0.496\) and \(0.504\), their Z-scores are calculated as -1.5 and 2.5, respectively.

These calculations indicate how many standard deviations the values of 0.496 and 0.504 are away from the mean of 0.499.
Deciphering Probability from Z-scores
Once the Z-score is determined, it can be crucial in calculating the probability of a particular value occurring within a normal distribution. Z-scores provide a standardized way to look at different data points in terms of their likelihoods.

To find these probabilities, a Z-table is typically used. A Z-table provides probabilities that a standard normal random variable is less than or equal to a given Z-score. For instance, in our exercise:
  • A Z-score of -1.5 corresponds to a probability of approximately 0.0668, meaning there’s a 6.68% chance that a bearing's diameter is below \(0.496\) inches.
  • The Z-score of 2.5 has a corresponding probability of approximately 0.9938, indicating a 99.38% chance that a diameter is below \(0.504\) inches.
By finding the difference between these probabilities, we can determine the likelihood of the diameter falling within the acceptable range, calculated as 0.9938 - 0.0668 = 0.927, or 92.7%.

This result indicates that 92.7% of the bearings are within the desired size range, while the remaining percentage falls outside, which is deemed unacceptable.
Explaining Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion in a set of data points. In a normal distribution, it tells us how spread out the numbers are from the mean. A small standard deviation means the values are closely grouped around the mean, whereas a large standard deviation indicates a wider spread.

In the context of our bearing production exercise, the standard deviation is 0.002 inches. This value quantifies the typical deviation each bearing's diameter might have from the average diameter of 0.499 inches.
  • The smaller the standard deviation, the more consistently the bearings are produced. This means less variability, and more bearings likely fall within the specified range of acceptable diameters.
  • Conversely, a larger standard deviation would suggest more variability in the bearing sizes, resulting in a more significant proportion falling outside the acceptable limits.
Understanding how standard deviation works provides crucial insights into the precision and quality of production processes, influencing decisions about adjustments needed to meet production standards.

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