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A manufacturing process is designed to produce bolts with a 0.5-in. diameter. Once each day, a random sample of 36 bolts is selected and the diameters recorded. If the resulting sample mean is less than \(0.49\) in. or greater than \(0.51\) in., the process is shut down for adjustment. The standard deviation for diameter is \(0.02\) in. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the true process mean really is \(0.5\) in.)

Short Answer

Expert verified
The probability of the manufacturing line being shut down unnecessarily is \(0.1336\) or \(13.36\% \).

Step by step solution

01

Define the Problem Variables

First, let's recognize the variables given in the problem - \(\mu = 0.5\), the true population mean. - \(\sigma = 0.02\), the true population standard deviation. - Sample size, \(n = 36\), the number of observations. We're looking for the probability that the sample mean \(\bar{x}\) falls outside the range \(0.49-0.51\).
02

Convert the Problem into a Standard Normal Distribution Problem

We're given a range where the production process is considered to be functioning correctly, \(0.49in < \bar{x} < 0.51in\). We must convert these to z-scores. To convert to z-scores, use the formula: \[Z = \frac{\bar{x} - \mu}{\frac{\sigma} {\sqrt{n}}}\] Where \(\bar{x} \) is the mean of our observations, \(\mu\) is the true mean, \(\sigma\) is the standard deviation, and \(n\) is the number of observations. In this case, transform the lower boundary (0.49in) and upper boundary (0.51in) to their corresponding z-values:
03

Compute the Lower Z-score:

Computation of lower z-score or z(lower boundary) provides: \[Z_{lower} = \frac{0.49 - 0.5}{\frac{0.02}{\sqrt{36}}} = -1.5\]
04

Compute the Upper Z-score:

Computation of upper z-score or z(upper boundary) provides: \[Z_{upper} = \frac{0.51 - 0.5}{\frac{0.02}{\sqrt{36}}} = 1.5\]
05

Calculate the Required Probability

We need to find the probability that the sample mean is outside this range, ie. \(P(Z < -1.5)\) or \(P(Z > 1.5)\). To find these, refer to the standard normal distribution table or calculator. The value corresponding to Z = 1.5 is \(0.4332\). Therefore, the probabilities we seek are : \[P(Z < -1.5) = P(Z > 1.5) = 0.5 - 0.4332\] Thus, the total probability of an unnecessary stoppage is given by two times the calculated probability.
06

Find the Result

Since we are looking for the manufacturing line to be shut down, either below the lower limit or above the upper limit. The probability of either event happening is calculated as: \[P = 2(0.5 - 0.4332) = 0.1336\]

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution that has a mean of 0 and a standard deviation of 1. It is a tool used in statistics to determine how data is dispersed around the mean, and it is fundamental to probability theory. When dealing with normal distributions that have different means and standard deviations, we can transform our measured values into z-scores to compare them to the standard normal distribution.

This conversion allows us to use standard normal distribution tables or calculators to find probabilities, which are essential for understanding and predicting outcomes in various fields, including manufacturing quality control. For example, the probability of a bolt’s diameter falling outside the acceptable range can be determined with the help of a standard normal distribution.
Sample Mean Calculation
The sample mean is an estimate of the population mean and is used to infer characteristics of the entire dataset from a smaller set of observations. To calculate the sample mean, you sum up all the observed values and then divide by the number of observations. In the context of manufacturing bolts, engineers take a sample to measure the mean diameter to decide whether the process needs adjustment. If the calculated sample mean diverges significantly from the expected process mean, it may indicate that something is amiss in the manufacturing process.

Accurate calculation of the sample mean is vital, as it allows quality control professionals to monitor process performance and take action when necessary to ensure that products meet specifications.
Z-score
A z-score, or standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It is calculated by subtracting the mean from the value in question and then dividing the result by the standard deviation.

In a manufacturing context, determining the z-score for a sample mean helps to identify how far off the sample mean is from the process mean. If the z-score is within a certain range, the process may be considered in control. If not, the process might need adjustment, as indicated in our bolt diameter example. A z-score gives professionals a clear, standardized method for flagging potential issues and maintaining production quality.
Statistical Process Control
Statistical process control (SPC) is a methodological framework used for monitoring and controlling a process to ensure that it operates at its fullest potential. One of the key tools of SPC is the control chart, which displays data over time and helps detect any significant variation from the mean.

In the case of bolt manufacturing, SPC would involve regularly sampling the diameters of bolts and plotting the sample means on a control chart. An SPC analysis might reveal whether the process variation is due to common causes (random variation inherent to the process) or special causes (variation due to specific, identifiable factors). Properly applied, SPC can prevent unnecessary adjustments to the manufacturing process and help ensure consistent product quality.

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Most popular questions from this chapter

For each of the following statements, identify the number that appears in boldface type as the value of either a population characteristic or a statistic: a. A department store reports that \(84 \%\) of all customers who use the store's credit plan pay their bills on time. b. A sample of 100 students at a large university had a mean age of \(24.1\) years. c. The Department of Motor Vehicles reports that \(22 \%\) of all vehicles registered in a particular state are imports. d. A hospital reports that based on the 10 most recent cases, the mean length of stay for surgical patients is \(6.4\) days. e. A consumer group, after testing 100 batteries of a certain brand, reported an average life of \(\mathbf{6 3} \mathrm{hr}\) of use.

A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large shipment is received, a random sample of 200 cartridges is selected, and each cartridge is inspected. If the sample proportion of defective cartridges is more than \(.02\), the entire shipment is returned to the vendor. a. What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is .05? b. What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is .10?

Explain the difference between a population characteristic and a statistic.

The nicotine content in a single cigarette of a particular brand has a distribution with mean \(0.8 \mathrm{mg}\) and standard deviation \(0.1 \mathrm{mg}\). If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than \(0.79 ?\) less than \(0.77\) ?

The thickness (in millimeters) of the coating applied to disk drives is a characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness \((x)\) has a normal distribution with a mean of \(3 \mathrm{~mm}\) and a standard deviation of \(0.05\) \(\mathrm{mm}\). Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining \(\bar{x}\), the mean coating thickness for the sample. a. Describe the sampling distribution of \(\bar{x}\) (for a sample of size 16 ). b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.) c. Referring to Part (b), what is the probability that a sample mean will be outside \(3 \pm 3 \sigma_{\bar{x}}\) just by chance (i.e., when there are no unusual circumstances)? d. Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of \(3.05 \mathrm{~mm}\). What is the probability that a problem will be detected when the next sample is taken? (Hint: This will occur if \(\bar{x}>3+3 \sigma_{\bar{x}}\) or \(\bar{x}<3-3 \sigma_{\bar{x}}\) when \(\mu=\) 3.05.) b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.)

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