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Weights of parts are normally distributed with variance \(\sigma^{2}\). Measurement error is normally distributed with mean zero and variance \(0.5 \sigma^{2}\), independent of the part weights, and adds to the part weight. Upper and lower specifications are centered at \(3 \sigma\) about the process mean. (a) Without measurement error, what is the probability that a part exceeds the specifications? (b) With measurement error, what is the probability that a part is measured as beyond specifications? Does this imply it is truly beyond specifications? (c) What is the probability that a part is measured beyond specifications if the true weight of the part is one \(\sigma\) below the upper specification limit?

Short Answer

Expert verified
(a) Probability is close to zero without error. (b) Low probability, but measurement may suggest being beyond specs falsely. (c) Probability is approximately 0.079.

Step by step solution

01

Understanding the Setup

The weights of parts have a normal distribution defined by variance \( \sigma^2 \). The specified limits are \( \pm 3\sigma \) from the mean. The measurement error follows a normal distribution with mean zero and variance \( 0.5 \sigma^2 \) and is independent of the actual weight of the parts.
02

Without Measurement Error

To determine the probability that a part exceeds the specifications without measurement error, consider the normal distribution centered at the process mean. Since the limits are \( \pm 3\sigma \), the probability of the weight exceeding these limits is practically zero due to the properties of a normal distribution (approximately 0.0027 in each tail region).
03

With Measurement Error

When measurement error is added, the part weight distribution has two components: the true weight and the error component. Both being normally distributed, their sum is also a normal distribution with a mean equal to the true weight mean and variance equal to \( 1.5\sigma^2 \) (combination of \( \sigma^2 \) and \( 0.5\sigma^2 \)). The specifications remain at \( \pm 3\sigma \). Using the normal distribution table or calculator, the probability of exceeding these limits with new variance will still be low since the measurement error variance is a small addition.
04

Probability with a Specific True Weight

When the part's true weight is one \( \sigma \) below the upper limit, calculate the probability that the weight measured is beyond the upper specification. The difference from the upper limit is \( \sigma \) with only the error variance to consider, i.e., \( 0.5\sigma^2 \). Calculate the probability of the error being greater than \( \sigma \), which corresponds to a standard normal deviation of \( \sqrt{2} \). Hence, the probability \( P(Z > \sqrt{2}) \) using the standard normal distribution is approximately 0.079.

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Key Concepts

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

Measurement Error
In engineering and manufacturing, precise measurements are crucial. However, measurement errors can happen, influencing the accuracy of results. Measurement error refers to the difference between a measured value and the actual value. In this exercise, the error is described by a normal distribution with a mean of zero and a variance of \(0.5\sigma^2\). This indicates that the error may alter the actual weight, but is expected to average out over time because it centers around zero.
  • **Importance:** Measurement errors can skew results, impacting quality control and decision-making.
  • **Characterization:** Random errors, like the ones here, are unavoidable but quantifiable, allowing engineers to consider them in evaluations.
  • **Impact:** Even though the variance is \(0.5\sigma^2\), it affects the final measurement's confidence by increasing the cumulative variance.
Managing measurement errors means recognizing their existence and factoring them into process evaluations to maintain product consistency. Understanding the variance helps in determining how much faith we can place in the measurements.
Process Specifications
Process specifications are boundaries set to ensure parts' quality. They define acceptable limits within which parts should fall. In this example, the specifications are established at \( \pm 3\sigma \) from the process mean.
The aim is to ensure that most parts conform to these limits, meaning only an extremely small fraction should fall out when everything operates correctly.
  • **Purpose:** They maintain product quality by categorizing something as within standard or not.
  • **Designing:** Located at \( \pm 3\sigma \), these are robust as approximately 99.73% of normally distributed data falls within this range, leaving only 0.27% outside if no error exists.
  • **Application:** Engineers use these to align processes so that parts consistently meet or exceed required quality standards.
Process specifications provide a clear quality benchmark, and placing them around three standard deviations from the mean is a common practice, ensuring high compliance and trust in the manufacturing process.
Probability Calculation
Calculating probabilities is crucial when dealing with part reliability and quality assurance. Here, understanding how likely an event is to occur helps us make informed decisions.Let's see how it applies to our parts' weights:
  • **Without Measurement Error:** The probability that a part's weight exceeds specifications is almost zero, as most data falls within \( \pm 3\sigma \). The remainder outside these limits is approximately 0.27%.
  • **With Measurement Error:** If the error has variance \(0.5\sigma^2\), it changes variance of overall weight measurements to \(1.5\sigma^2\). Though it slightly raises the probability of crossing limits, it remains low due to the small error variance.
  • **Specific Cases:** When a part's true weight is one \( \sigma \) below the upper limit, the probability of exceeding this boundary is calculated using the error variance, demonstrating the precise impact of tiny shifts due to error.
Probability calculations in engineering often involve determining these chances under varying conditions to ensure that components meet stringent specifications. Thorough assessments help prevent deviations and inform adjustments needed to align processes.

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Most popular questions from this chapter

If \(X\) and \(Y\) have a bivariate normal distribution with joint probability density \(f_{X Y}\left(x, y ; \sigma_{X}, \sigma_{Y}, \mu_{X}, \mu_{Y}, \rho\right),\) show that the marginal probability distribution of \(X\) is normal with mean \(\mu_{X}\) and standard deviation \(\sigma_{X}\). [Hint: Complete the square in the exponent and use the fact that the integral of a normal probability density function for a single variable is \(1 .]\)

The yield in pounds from a day's production is normally distributed with a mean of 1500 pounds and standard deviation of 100 pounds. Assume that the yields on different days are independent random variables. (a) What is the probability that the production yield exceeds 1400 pounds on each of five days next week? (b) What is the probability that the production yield exceeds 1400 pounds on at least four of the five days next week?

A plastic casing for a magnetic disk is composed of two halves. The thickness of each half is normally distributed with a mean of 2 millimeters and a standard deviation of 0.1 millimeter and the halves are independent. (a) Determine the mean and standard deviation of the total thickness of the two halves. (b) What is the probability that the total thickness exceeds 4.3 millimeters?

Soft-drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables. (a) What is the standard deviation of the average fill volume of 100 cans? (b) If the mean fill volume is 12.1 ounces, what is the probability that the average fill volume of the 100 cans is below 12 fluid ounces? (c) What should the mean fill volume equal so that the probability that the average of 100 cans is below 12 fluid ounces is \(0.005 ?\) (d) If the mean fill volume is 12.1 fluid ounces, what should the standard deviation of fill volume equal so that the probability that the average of 100 cans is below 12 fluid ounces is \(0.005 ?\) (e) Determine the number of cans that need to be measured such that the probability that the average fill volume is less than 12 fluid ounces is 0.01 .

Making handcrafted pottery generally takes two major steps: wheel throwing and firing. The time of wheel throwing and the time of firing are normally distributed random variables with means of \(40 \mathrm{~min}\) and \(60 \mathrm{~min}\) and standard deviations of 2 min and 3 min, respectively. (a) What is the probability that a piece of pottery will be finished within 95 min? (b) What is the probability that it will take longer than \(110 \mathrm{~min} ?\)

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