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Suppose that a quality characteristic is normally distributed with specifications from 10 to 30 units. The process standard deviation is 2 units. (a) Calculate the natural tolerance limits, fraction defective, \(P C R,\) and \(P C R_{k}\) when the process mean is \(20 .\) (b) Suppose that the process mean shifts higher by 1.5 standard deviations. Recalculate the quantities in part (b). (c) Compare the results in parts (a) and (b) and comment on any differences.

Short Answer

Expert verified
Initial centered mean: PCR and PCRk = 1.67, no defectives. Shifted mean: FRACTION DEFECTIVES ~0.00023, PCRk reduced to 1.17.

Step by step solution

01

Calculate Natural Tolerance Limits

The process has a normal distribution with a mean of 20 and a standard deviation of 2. The natural tolerance limits are calculated using the formula: \[ \text{Natural tolerance limits} = \mu \pm 3\sigma \]where \(\mu = 20\) and \(\sigma = 2\).Calculating:\[ 20 - 3(2) = 14 \quad \text{and} \quad 20 + 3(2) = 26 \]So, the natural tolerance limits are 14 to 26 units.
02

Determine Fraction Defective

The specification limits are from 10 to 30 units. To determine the fraction defective, we need to calculate the cumulative probabilities outside the specification limits.The Z-scores for the specification limits of 10 and 30 are:\[ Z_{10} = \frac{10 - 20}{2} = -5 \]\[ Z_{30} = \frac{30 - 20}{2} = 5 \]Using Z-tables (or standard normal distribution table), find \(P(Z < -5)\) and \(P(Z > 5)\); both are almost 0. Therefore, the fraction of defectives is approximately 0.
03

Calculate Process Capability Ratio (PCR)

The process capability ratio \(PCR\) is given by the formula:\[ PCR = \frac{USL - LSL}{6\sigma} \]where USL = 30, LSL = 10, and \(\sigma = 2\).Plug in the values:\[ PCR = \frac{30 - 10}{6(2)} = \frac{20}{12} = \frac{5}{3} \approx 1.67 \]
04

Calculate Centered Process Capability Ratio (PCRk)

The centered process capability ratio \(PCR_k\) is calculated as:\[ PCR_k = \frac{min(\mu - LSL, USL - \mu)}{3\sigma} \]Calculating both differences: \[ \mu - 10 = 20 - 10 = 10 \]\[ 30 - \mu = 30 - 20 = 10 \]Since both values are equal, \[ PCR_k = \frac{10}{6} \approx 1.67 \]
05

Adjust Process Mean for Shift

Now assume that the process mean shifts higher by 1.5 standard deviations. The new mean \(\mu'\) is:\[ \mu' = 20 + 1.5(2) = 23 \]
06

Recalculate New Natural Tolerance Limits

With the new mean of 23:\[ \text{New Natural Tolerance Limits} = 23 \pm 3(2) = 17 \quad \text{to} \quad 29 \]
07

Recalculate Fraction Defective with Shifted Mean

With the new mean of 23:Find the Z-scores for the specification limits:\[ Z_{10}' = \frac{10 - 23}{2} = -6.5 \]\[ Z_{30}' = \frac{30 - 23}{2} = 3.5 \]Using Z-tables, \(P(Z < -6.5) \approx 0\) and \(P(Z > 3.5) \approx 0.00023\). Therefore, the fraction defective \( \approx 0.00023\).
08

Recalculate Process Capability Ratios with Shifted Mean

For \(PCR\):The formula remains the same:\[ \frac{30 - 10}{6(2)} = 1.67 \]For \(PCR_k\):\[ \mu' - LSL = 23 - 10 = 13 \]\[ USL - \mu' = 30 - 23 = 7 \]\[ PCR_k = \frac{7}{6} \approx 1.17 \]
09

Comparison between Initial and Shifted Mean Results

In part (a), the process mean was centered, providing a balance and high \(PCR_k\). In part (b), the mean shift led to an increase in defective units and a drop in \(PCR_k\) to 1.17, indicating reduced capability to produce within specs when the process is not centered.

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Key Concepts

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

Normal Distribution
Imagine you have a process that creates items and want to understand how they are spread or varied around a central value. That's where the Normal Distribution comes in handy. It's a bell-shaped curve that helps to visualize the spread and concentrates most values around the mean. For quality control, it keeps track of variation and predicts process behaviors. Often used in quality control, assuming your process follows normal distribution gives insight into performance by letting you calculate probabilities and predict outcomes.

In our case, imagine a process where item characteristics are normally distributed with a mean of 20 and a standard deviation of 2 units. This distribution allows us to calculate how much data falls within a specific range and is fundamental to controlling product quality. The natural tolerance limits, for instance, span from 14 to 26, showing where most of your product measurements should fall if your process is well-controlled.
Quality Control
Quality Control ensures products or services meet specific quality standards consistently. It evaluates whether the process stays within set limits. The main goal is to minimize defects and maximize customer satisfaction. Methods like Process Capability Analysis and Statistical Process Control (SPC) are instrumental here.

By adopting these methodologies, companies can monitor and control their production processes. Quality control revolves around maintaining stable processes and using statistical tools to find inconsistencies and errors. This proactive approach leads to fewer defects and enhances production consistency.
Statistical Process Control (SPC)
Statistical Process Control (SPC) is a method used to monitor and control a process through statistical means. By employing SPC, you can determine whether a process is stable and predictable over time. It involves collecting data over time and then using this data to identify trends or variations that occur in the process.

SPC is all about understanding variation—one of the key principles to maintaining efficient quality control. It uses control charts, a core tool for understanding the variability in your process and signals when it needs improvement. When variations exceed the expected or control limits, it can signal an issue necessitating corrective measures to bring the process back to a more consistent state.
Z-scores
Z-scores play a crucial role in understanding where a specific value lies within a normal distribution. Defined as the number of standard deviations away a point is from the mean, Z-scores help determine how unusual or typical a value is.

For example, when calculating the Z-score for specification limits in our scenario, we found it to be either -5 or 5 initially, which showed an almost zero probability for defects, emphasizing that the process was well within limits. Z-scores help convert your process measurements into standardized scores, making it easier to decide if your process is operating correctly or if adjustments are necessary.
Process Mean Shift
The Process Mean Shift happens when the average (mean) of your process data moves over time. Such shifts can occur from various factors, including changes in materials, environment, equipment, or how the process is managed.

In our example, the mean shifted from 20 to 23 when it moved by 1.5 standard deviations. This shift significantly affects process capability, as seen by a decrease in the centered process capability ratio (PCR_k). A mean shift might signal potential issues in consistency and increase the likelihood of producing defectives. Monitoring for mean shifts is crucial for maintaining control and ensuring that the process delivers what is expected.

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Most popular questions from this chapter

An article in Quality \& Safety in Health Care ["Statistical Process Control as a Tool for Research and Healthcare Improvement," (2003) Vol. \(12,\) pp. \(458-464]\) considered a number of control charts in healthcare. The following approximate data were used to construct \(\bar{X}-S\) charts for the turn around time (TAT) for complete blood counts (in minutes). The subgroup size is \(n=3\) per shift, and the mean standard deviation is \(21 .\) Construct the \(\bar{X}\) chart and comment $$\begin{array}{cc|c|c|c|c|c|c|c|c|c|c|c|c|c|}t & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\\\\hline \text { TAT } & 51 & 73 & 28 & 52 & 65 & 49 & 51 & 50 & 25 & 39 & 40 & 30 & 49 & 31 \\\\\hline\end{array}$$ on the control of the process. If necessary, assume that assignable causes can be found, eliminate suspect points, and revise the control limits.

An article in Electric Power Systems Research ["On the Self-Scheduling of a Power Producer in Uncertain Trading Environments" \((2008,78(3),\) pp. \(311-317)]\) considered a selfscheduling approach for a power producer. The following table shows the forecasted prices of energy for a 24 -hour time period according to a base case scenario. $$\begin{array}{cccccc}\hline \text { Hour } & \text { Price } & \text { Hour } & \text { Price } & \text { Hour } & \text { Price } \\\\\hline 1 & 38.77 & 9 & 48.75 & 17 & 52.07 \\\2 & 37.52 & 10 & 51.18 & 18 & 51.34 \\\3 & 37.07 & 11 & 51.79 & 19 & 52.55 \\\4 & 35.82 & 12 & 55.22 & 20 & 53.11 \\\5 & 35.04 & 13 & 53.48 & 21 & 50.88 \\\6 & 35.57 & 14 & 51.34 & 22 & 52.78 \\\7 & 36.23 & 15 & 45.8 & 23 & 42.16 \\\8 & 38.93 & 16 & 48.14 & 24 & 42.16\end{array}$$ \ (a) Construct individuals and moving-range charts. Determine whether the energy prices fluctuate in statistical control. (b) Is the assumption of independent observations reasonable for these data?

The following data are the number of spelling errors detected for every 1000 words on a news Web site over 20 weeks. $$\begin{array}{cccc}\hline \text { Week } & \begin{array}{l}\text { No. of Spelling } \\ \text { Errors }\end{array} & \text { Week } & \begin{array}{l}\text { No. of Spelling } \\\\\text { Errors }\end{array} \\\\\hline 1 & 3 & 11 & 1 \\\2 & 6 & 12 & 6 \\\3 & 0 & 13 & 9 \\\4 & 5 & 14 & 8 \\\5 & 9 & 15 & 6 \\\6 & 5 & 16 & 4 \\\7 & 2 & 17 & 13 \\\8 & 2 & 18 & 3 \\\9 & 3 & 19 & 0 \\\10 & 2 & 20 & 7 \\\\\hline\end{array}$$ (a) What control chart is most appropriate for these data? (b) Using all the data, compute trial control limits for the chart in part (a), construct the chart, and plot the data. (c) Determine whether the process is in statistical control. If not, assume that assignable causes can be found and outof-control points eliminated. Revise the control limits.

A control chart for individual observations has 3 -sigma control limits \(U C L=1.80\) and \(L C L=1.62 .\) The process specification limits are (1.64,1.84) . (a) Estimate the process standard deviation. (b) Calculate \(P C R\) and \(P C R_{k}\) for the process.

In a semiconductor manufacturing process, CVD metal thickness was measured on 30 wafers obtained over approximately two weeks. Data are shown in the following table. (a) Using all the data, compute trial control limits for individual observations and moving-range charts. Construct the chart and plot the data. Determine whether the process is in statistical control. If not, assume that assignable causes can be found to eliminate these samples and revise the control limits. (b) Estimate the process mean and standard deviation for the in-control process. $$\begin{array}{cccc}\hline \text { Wafer } & {x} & \text { Wafer } & {x} \\\\\hline 1 & 16.8 & 16 & 15.4 \\\2 & 14.9 & 17 & 14.3 \\\3 & 18.3 & 18 & 16.1 \\\4 & 16.5 & 19 & 15.8 \\\5 & 17.1 & 20 & 15.9 \\\6 & 17.4 & 21 & 15.2 \\\7 & 15.9 & 22 & 16.7 \\\8 & 14.4 & 23 & 15.2 \\\9 & 15.0 & 24 & 14.7 \\\10 & 15.7 & 25 & 17.9 \\\11 & 17.1 & 26 & 14.8 \\\12 & 15.9 & 27 & 17.0 \\\13 & 16.4 & 28 & 16.2 \\\14 & 15.8 & 29 & 15.6 \\\15 & 15.4 & 30 & 16.3\end{array}$$

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