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Chapter 7: Problem 19

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
3.74% of slots do not meet the specifications.

Step by step solution

01

Understand the Problem

We have a Normal distribution with mean \( \mu = 0.8750 \) and standard deviation \( \sigma = 0.0012 \). We need to find the proportion of slot widths that fall outside the range of 0.8725 to 0.8775 inches.
02

Find the Z-scores

The Z-score is calculated by \( Z = \frac{X - \mu}{\sigma} \). For the specification limits of 0.8725 and 0.8775:1. Lower limit: \( Z_1 = \frac{0.8725 - 0.8750}{0.0012} = \frac{-0.0025}{0.0012} = -2.0833 \).2. Upper limit: \( Z_2 = \frac{0.8775 - 0.8750}{0.0012} = \frac{0.0025}{0.0012} = 2.0833 \).
03

Use the Standard Normal Table

Look up the Z-scores in a standard Normal distribution table to find the probabilities:1. For \( Z_1 = -2.0833 \), the probability \( P(Z < -2.0833) \) is approximately 0.0187.2. For \( Z_2 = 2.0833 \), the probability \( P(Z < 2.0833) \) is approximately 0.9813.
04

Calculate the Proportion Outside the Range

The proportion of widths that fall within the range is given by \( P(-2.0833 < Z < 2.0833) = P(Z < 2.0833) - P(Z < -2.0833) = 0.9813 - 0.0187 = 0.9626 \).
05

Find the Complement to Determine Non-conformance

The proportion of widths that do not meet specifications is the complement of the proportion within specifications, calculated as follows: \( 1 - 0.9626 = 0.0374 \).

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Key Concepts

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

Z-score
To tackle problems involving the normal distribution, we often use the Z-score. The Z-score is a measurement that describes a value's position in relation to the mean of a group of values. This helps standardize different data points for easy comparison. To compute a Z-score, you subtract the mean from the data point you are interested in and then divide that result by the standard deviation. The equation is given by: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the value we are examining,
  • \(\mu\) is the mean of the data, and
  • \(\sigma\) is the standard deviation.
This calculation centers the data around a mean of zero and a standard deviation of one. In our example, the Z-scores are \(-2.0833\) for the lower limit and \(2.0833\) for the upper limit. This tells us how far each boundary is from the mean, in terms of standard deviations. The farther away the Z-score from zero, the more unusual the data point is in the distribution.
Standard Normal Table
Once we find Z-scores, the next step is to use the Standard Normal Table, also known as the Z-table. This table helps us determine the area under the standard normal curve to the left of a given Z-score. This area represents the probability that a value will fall below the specified Z-score in a standard normal distribution.When using a Z-table:
  • Locate the row that contains the integer and first decimal place of the Z-score.
  • Cross-reference it with the column that contains the second decimal place of the Z-score.
In our case, a Z-score of \(-2.0833\) corresponds to a probability of approximately 0.0187, indicating the chance of a slot width being less than 0.8725 inches. Meanwhile, a Z-score of \(2.0833\) corresponds to a probability of 0.9813 for widths less than 0.8775 inches. By examining these probabilities, we understand the distribution of values about the mean.
Proportion Calculation
After finding the Z-scores and corresponding probabilities using the Z-table, the next step is calculating the proportion of the data that falls within or outside a specified range. For our exercise, we want to find the proportion of slots not meeting specified width limits.To find this, follow these steps:
  • Calculate the proportion of data within the specified range. This involves finding the difference between the probability of the upper boundary and the lower boundary: \( P(-2.0833 < Z < 2.0833) = P(Z < 2.0833) - P(Z < -2.0833) = 0.9813 - 0.0187 = 0.9626 \).
  • Determine the proportion of data falling outside the range by subtracting the within-range proportion from 1. This is the complement and is given by: \( 1 - 0.9626 = 0.0374 \).
Thus, about 3.74% of slot widths will not meet the specifications, given the normal distribution of the data. Understanding these calculations helps us quantitatively assess how conforming or non-conforming our dataset is relative to specified conditions.

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Most popular questions from this chapter

What are the most important differences between female and male students? (a) A greater percentage of females spend five or more hours per day playing video or computer games or using the computer for something that is not school work, on an average school day, than males. (b) Females are more than twice as likely to spend no time playing video or computer games or using the computer for something that is not school work, on an average school day, than males. (c) A greater percentage of females spend one hour or less per day playing video or computer games or using the computer for something that is not school work, on an average school day, than males. (d) All of the above.

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