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Chapter 1: Problem 29

A machine produces bearings with mean diameter 3.00 inches and standard deviation 0.01 inch. Bearings with diameters in excess of 3.02 inches or less than 2.98 inches will fail to meet quality specifications. a. Approximately what fraction of this machine's production will fail to meet specifications? b. What assumptions did you make concerning the distribution of bearing diameters in order to answer this question?

Short Answer

Expert verified
Approximately 4.56% of bearings will fail specifications, assuming a normal distribution.

Step by step solution

01

Understanding the Problem

We need to find the fraction of bearings that are produced with diameters outside the specification limits. The acceptable range for the diameter is between 2.98 inches and 3.02 inches. The mean diameter is 3.00 inches with a standard deviation of 0.01 inches.
02

Determine the Distribution Assumptions

We assume the diameters of the bearings are normally distributed because this is a common assumption when dealing with natural variations in manufacturing processes. This allows us to use the properties of the normal distribution to find probabilities.
03

Calculate the Z-scores for Specification Limits

Calculate the Z-scores for the limits of 2.98 inches and 3.02 inches. Use the formula: \[ Z = \frac{X - \mu}{\sigma} \]For 2.98 inches: \[ Z_{lower} = \frac{2.98 - 3.00}{0.01} = -2.00 \]For 3.02 inches: \[ Z_{upper} = \frac{3.02 - 3.00}{0.01} = 2.00 \]
04

Find the Probability for Each Z-score

Using the standard normal distribution table, find the probability corresponding to the Z-scores. For \( Z = -2.00 \), the probability is approximately 0.0228. For \( Z = 2.00 \), the probability is approximately 0.9772.
05

Calculate the Percentage Outside the Limits

To find the fraction of production that fails to meet specifications, calculate the total area outside the acceptable range by subtracting the probabilities obtained in Step 4: \[ P(X < 2.98) + P(X > 3.02) = 0.0228 + (1 - 0.9772) = 0.0228 + 0.0228 = 0.0456 \] Thus, approximately 4.56% of the bearings will fail to meet specifications.

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Key Concepts

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

Standard Deviation
The standard deviation is a measure of how spread out numbers are in a data set. In the context of the normal distribution, it tells us how much individual observations in a group differ from the mean. When we say that the standard deviation of bearing diameters is 0.01 inches, this implies that most bearings produced by the machine will have diameters within 0.01 inches of the average diameter of 3.00 inches.

A smaller standard deviation indicates that the diameters are tightly clustered around the mean, while a larger standard deviation would mean the diameters are more spread out. To better understand the use of standard deviation:
  • Helps to gauge consistency in manufacturing, a smaller value means more consistent output.
  • In a normal distribution, about 68% of data points lie within one standard deviation of the mean, 95% within two, and 99.7% within three.
Understanding standard deviation is crucial as it plays a vital role in measuring reliability and precision in processes like manufacturing.
Z-scores
Z-scores are a statistical measurement that indicates how many standard deviations an element is from the mean. They are essential when assessing the likelihood of a data point occurring under a normal distribution. In this scenario, Z-scores help determine how far each limit (2.98 inches and 3.02 inches) is from the mean diameter of 3.00 inches.

The formula for calculating Z-scores is:\[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the data point.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
For bearing diameters:
  • The Z-score for 2.98 inches is \(-2.00\).
  • The Z-score for 3.02 inches is \(2.00\).
These Z-scores indicate how the bearing diameters align with standard deviation units. Positive Z-scores denote values above the mean, while negative ones are below it. Z-scores allow us to use standard normal distribution tables effectively to calculate probabilities.
Probability
Probability in the context of normal distribution assesses the likelihood of random outcomes. Here, we're interested in the probability of bearings falling outside the specification limits.

We use Z-scores to find these probabilities through a standard normal distribution table:
  • For a Z-score of -2.00, the table provides a probability of approximately 0.0228, which indicates the likelihood of a bearing being below 2.98 inches.
  • For a Z-score of 2.00, the probability is around 0.9772, meaning the probability of a bearing being above this value is \(1 - 0.9772 = 0.0228\).
By adding these two probabilities, we find the overall chance of a bearing being outside the specification limits:\[ P(X < 2.98) + P(X > 3.02) = 0.0228 + 0.0228 = 0.0456 \]Thus, approximately 4.56% of the bearings are expected to be out of the desired range. This aspect of probability allows manufacturers to assess quality control and make adjustments to maintain high standards.

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Most popular questions from this chapter

Of great importance to residents of central Florida is the amount of radioactive material present in the soil of reclaimed phosphate mining areas. Measurements of the amount of \(^{238} \mathrm{U}\) in 25 soil samples were as follows (measurements in picocuriespergram):$$\begin{array}{rrrrr}.74 & 6.47 & 1.90 & 2.69 & .75 \\\\.32 & 9.99 & 1.77 & 2.41 & 1.96 \\\1.66 & .70 & 2.42 & .54 &3.36 \\\3.59 & .37 & 1.09 & 8.32 & 4.06 \\ 4.55 & .76 & 2.03 & 5.70 & 12.48\end{array}$$.Construct a relative frequency histogram for these data.

Over the past year, a fertilizer production process has shown an average daily yield of 60 tons with a variance in daily yields of \(100 .\) If the yield should fall to less than 40 tons tomorrow, should this result cause you to suspect an abnormality in the process? (Calculate the probability of obtaining less than 40 tons. What assumptions did you make concerning the distribution of yields?

Let \(k \geq 1\). Show that, for any set of \(n\) measurements, the fraction included in the interval \(\bar{y}-k s\) to \(\bar{y}+k s\) is at least \(\left(1-1 / k^{2}\right) .\) [Hint: $$s^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}\right]$$ In this expression, replace all deviations for which \(\left|y_{i}-\bar{y}\right| \geq k s\) with \(k s\). Simplify.] This result is known as Tchebysheff's theorem. \(^{*}\)

The discharge of suspended solids from a phosphate mine is normally distributed with mean daily discharge 27 milligrams per liter \((\mathrm{mg} / \mathrm{L})\) and standard deviation \(14 \mathrm{mg} / \mathrm{L}\). In what proportion of the days will the daily discharge be less than \(13 \mathrm{mg} / \mathrm{L} ?\)

The following results on summations will help us in calculating the sample variance \(s^{2}\). For any constant \(c\), a. \(\sum_{i=1}^{n} c=n c\). b. \(\sum_{i=1}^{n} c y_{i}=c \sum_{i=1}^{n} y_{i}\). c. \(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} y_{i}\). Use (a), (b), and (c) to show that $$s^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n} y_{i}^{2}-\frac{1}{n}\left(\sum_{i=1}^{n} y_{i}\right)^{2}\right].$$
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