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Automated manufacturing operations are quite precise but still vary, often with distributions that are close to Normal. The width, in inches, of slots cut by a milling machine follows approximately the \(N(0.8750,0.0012)\) distribution. The specifications allow slot widths between \(0.8725\) and \(0.8775\) inch. What proportion of slots do not meet these specifications?

Short Answer

Expert verified
3.76% of slots do not meet the specifications.

Step by step solution

01

Understanding the Problem

The slot width is a normally distributed variable with a mean of 0.8750 inches and a standard deviation of 0.0012 inches. We need to find the proportion of slots that fall outside the specification limits of 0.8725 inches to 0.8775 inches.
02

Calculating Z-scores for Specification Limits

The Z-score for a normal distribution is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \]For the lower limit (0.8725), the Z-score is \[ Z_{lower} = \frac{0.8725 - 0.8750}{0.0012} = -2.08 \]For the upper limit (0.8775), the Z-score is \[ Z_{upper} = \frac{0.8775 - 0.8750}{0.0012} = 2.08 \]
03

Finding the Proportion within the Specification Limits

We use the standard normal distribution table to find the area between these Z-scores. The probability for a Z-score of -2.08 is approximately 0.0188, and for 2.08 is 0.9812. Thus, the proportion of slots within the limits is \[ P(-2.08 < Z < 2.08) = 0.9812 - 0.0188 = 0.9624 \]
04

Calculating the Proportion Outside Specification Limits

To find the proportion of slots outside the specification limits, subtract the proportion within the limits from 1: \[ P = 1 - 0.9624 = 0.0376 \]
05

Conclusion

The proportion of slots that do not meet the specifications is 0.0376.

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Key Concepts

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

Z-scores
Z-scores are a handy way to understand where a specific data point stands in a distribution. Think of Z-scores as a ruler that measures how far away a point is from the mean, in terms of standard deviations. For example, if we have a set of data that is normally distributed (like the slot widths in this exercise), Z-scores help us figure out how extreme or typical a data value is compared to the average value. This is important because it translates raw scores into a format that helps us make probabilistic conclusions.The formula for finding a Z-score is quite simple: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • X is the value of the data point you are examining,
  • \( \mu \) (mu) is the mean, or average, of the distribution, and
  • \( \sigma \) (sigma) is the standard deviation of the distribution.
Z-scores can be positive or negative:
  • A positive Z-score means the data point is above the mean,
  • while a negative Z-score means it's below the mean.
In the exercise, Z-scores were calculated for the specification limits to determine how far these limits are from the mean slot width. This helps in finding the probability of slot widths lying outside these limits.
Standard Deviation
Standard deviation is like a guide that tells us how much the values in a data set deviate, or vary, from the mean. In simple terms, it measures the spread or dispersion of a set of data. When the standard deviation is low, data points are close to the mean, and when it's high, data points are spread out over a wider range. In a normal distribution, about 68% of the data falls within one standard deviation of the mean, around 95% within two standard deviations, and 99.7% within three. This is known as the Empirical Rule. In this exercise, the slot widths have a small standard deviation of 0.0012 inches, indicating that most of the slots will have widths quite close to the mean of 0.8750 inches. The standard deviation is crucial when calculating Z-scores because it scales how far our "specification limit" data points are from the mean, which allows us to understand the probability of these limits being exceeded.
Probability Calculation
Probability calculation in the context of normal distribution revolves around finding the likelihood of specific events occurring. Here, it involves determining how likely it is for slot widths to fall within or outside the specified limits using Z-scores and a standard normal distribution table. Once Z-scores for the upper and lower specification limits are calculated, we consult the standard normal distribution table, which tells us the probability of a data point being below certain Z-scores. The exercise used these probabilities to compute the proportion of slots within the specification limits as follows: - Find the probability for each Z-score using the distribution table. - For the lower Z-score of -2.08, the probability was 0.0188, and for the upper Z-score of 2.08, it was 0.9812. By subtracting these probabilities, we find the probability of slots being within these limits. Thus, most slots are within specifications, confirming the precise nature of manufacturing in this context.
Specification Limits
Specification limits are like the boundaries within which the data must fall to meet acceptable standards. They are crucial in quality control, especially in manufacturing processes like the milling machine in our exercise, which must produce slot widths within tightly defined limits of 0.8725 and 0.8775 inches. When measurements fall outside of these limits, it indicates potential issues in the manufacturing process, which could lead to defective products. In statistical terms, we call such deviations outliers, which are undesirable in precision-focused tasks. In the given problem, specification limits help to decide the acceptable range of slot widths. By calculating the Z-scores for these limits, we can efficiently determine the probability and thus the proportion of slots that fall outside the desirable range. This process is a fundamental aspect of maintaining and assessing quality in industrial processes, helping to ensure products meet necessary standards before reaching the customer.

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Most popular questions from this chapter

Reports on a student's ACT, SAT, or MCAT usually give the percentile as well as the actual score. The percentile is the cumulative proportion stated as a percentage: the percentage of all scores that were lower than this one. In 2019 , the scores on the Mathematics portion of the SAT were close to Normal, with mean 528 and standard deviation 117. 5 a. Find the 85 th percentile for the scores on the Mathematics portion of the SAT. b. Joseph scored 451 . What was his percentile? c. Find the first quartile for the scores on the Mathematics portion of the SAT.

A report gives the mean and median credit card debt per American household as of October 2019. The two values that are reported are \(\$ 2300\) and \(\$ 5700\). Which of these is the mean? Explain how you know this.

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