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Suppose that a quality characteristic is normally distributed with specifications at \(100 \pm 20 .\) The process standard deviation is 6 . (a) Suppose that the process mean is \(100 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{k}\) and interpret these ratios. (b) Suppose that the process mean is \(106 .\) What are the natural tolerance limits? What is the fraction defective? Calculate \(P C R\) and \(P C R_{c}\) and interpret these ratios.

Short Answer

Expert verified
For part (a), both PCR and PCR_k are 1.11 with negligible defective fraction. For part (b), PCR is 1.11, PCR_c is 0.78, and about 0.99% defective due to upper mean shift.

Step by step solution

01

Find Natural Tolerance Limits

For a normally distributed process with mean \( \mu \) and process standard deviation \( \sigma \), the natural tolerance limits are given by \( \mu \pm 3\sigma \). For part (a) with mean \( \mu = 100 \) and \( \sigma = 6 \), the natural tolerance limits are \( 100 - 18 = 82 \) and \( 100 + 18 = 118 \).
02

Calculate Fraction Defective

The fraction defective is the probability that a measurement falls outside the specification limits.The specification limits are \( 100 - 20 = 80 \) and \( 100 + 20 = 120 \). Since the natural tolerance limits \( (82, 118) \) fall within the specification limits \( (80, 120) \), the fraction defective is negligible, essentially close to zero.
03

Calculate PCR

The process capability ratio, \( PCR \), is calculated as:\[PCR = \frac{USL - LSL}{6 \sigma}\]Substitute \( USL = 120 \), \( LSL = 80 \), and \( \sigma = 6 \):\[PCR = \frac{120 - 80}{36} = \frac{40}{36} \approx 1.11\]
04

Calculate PCR_k

The process capability index, \( PCR_k \), accounts for any shift in the process mean and is calculated as:\[PCR_k = \frac{1}{3} \min\left(\frac{USL - \mu}{\sigma}, \frac{\mu - LSL}{\sigma}\right)\]With \( \mu = 100 \), \( \sigma = 6 \):\[PCR_k = \frac{1}{3} \min\left(\frac{120 - 100}{6}, \frac{100 - 80}{6}\right) = \frac{1}{3} \min\left(3.33, 3.33\right) = 1.11\]
05

Part (b) - Adjust Mean to 106 and Find New Natural Tolerance Limits

For part (b), change \( \mu \) to 106.The new natural tolerance limits are \( 106 - 18 = 88 \) and \( 106 + 18 = 124 \).
06

Calculate Fraction Defective for New Mean

For the new mean, since the upper natural tolerance of 124 exceeds the upper specification limit of 120, calculate the fraction defective for this upper end.The fraction defective is the probability that a measurement exceeds 120, use the standard normal distribution \( Z = \frac{X - \mu}{\sigma} \):\[Z = \frac{120 - 106}{6} = 2.33\]Evoking standard normal tables or software, \( P(Z > 2.33) \approx 0.0099 \), so about 0.99% defective.
07

Calculate New PCR

The process capability ratio, \( PCR \), is the same calculation as before, as the specification limits and \( \sigma \) do not change.So, \( PCR = 1.11 \).
08

Calculate PCR_c

The process capability index, \( PCR_c \), for the mean of 106 and considering mean shift, is:\[PCR_c = \frac{1}{3} \min\left(\frac{120 - 106}{6}, \frac{106 - 80}{6}\right) = \frac{1}{3} \min\left(2.33, 4.33\right) = 0.78\]
09

Interpret PCR and PCR_k

For part (a), \( PCR = 1.11 \) and \( PCR_k = 1.11 \), suggest that the process is generally capable and centered.For part (b), \( PCR = 1.11 \) but \( PCR_c = 0.78 \) indicates that although the process can meet specifications, it is not well-centered, with potential higher defect rates at extremes.

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Key Concepts

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

Natural Tolerance Limits
Natural Tolerance Limits (NTLs) are a range that describes the natural spread of a process. This range is critical to understanding the variability in quality characteristics. For a process that follows a normal distribution, NTLs are determined by the formula: \[ \mu \pm 3\sigma \]Where \( \mu \) is the process mean, and \( \sigma \) is the standard deviation. The limits hence cover 99.73% of the process outputs. It's like drawing a boundary within which most of the data points will fall under normal operations. In our exercise, when the process mean is 100 and \( \sigma = 6 \), the NTLs are calculated as:* Lower Limit: \( 100 - 18 = 82 \)* Upper Limit: \( 100 + 18 = 118 \)These calculations are crucial because they help in predicting whether a process is capable of meeting set specifications, such as \( 100 \pm 20 \). By comparing NTLs with specification limits, we get a clear idea of the process's natural variability and how likely it is to produce defects.
Fraction Defective
Fraction Defective refers to the proportion of products or outputs that fall outside the specified limits. Essentially, it shows how many units are not meeting the quality standards and may require rework or rejection. To find the fraction defective, we consider the specification limits and assess where the natural tolerance limits fall in relation to these. If NTLs lie within the specification limits, the fraction defective can be negligible. For example, in part (a) of our exercise, the specification limits are 80 to 120, and the NTLs (82 to 118) fall within these bounds, meaning the fraction defective is very low, almost zero. For part (b), with the mean adjusted to 106, the new NTLs are 88 to 124. In this case, the upper NTL exceeds the upper specification limit (120), indicating a potential for outputs to exceed quality expectations. Here, the fraction defective is computed using a Z-score: \[ Z = \frac{120 - 106}{6} = 2.33 \]The probability of Z exceeding 2.33 can be found using standard normal distribution tables or software, leading to an approximate defect rate of 0.99%.
Process Capability Ratio (PCR)
The Process Capability Ratio (PCR) is a simple yet powerful way to assess a process's ability to produce outputs within specified limits. It is defined as:\[ PCR = \frac{USL - LSL}{6 \sigma} \]Where \( USL \) is the Upper Specification Limit, \( LSL \) is the Lower Specification Limit, and \( \sigma \) is the standard deviation. This ratio effectively measures how well the 6-sigma range fits within the specification intervals. In the exercise, using \( USL = 120 \), \( LSL = 80 \), and \( \sigma = 6 \), we compute:\[ PCR = \frac{40}{36} \approx 1.11 \]A PCR greater than 1 suggests that the process is generally capable of staying within limits, as there is a buffer between process variation and specification extremes. However, PCR doesn't account for whether the process mean is centered or shifted towards one limit, which could result in a different interpretation if not checked alongside other metrics.
Process Capability Index (PCR_k)
Process Capability Index (PCR_k), sometimes denoted as Cpk, offers a more comprehensive insight into process capability by measuring not only the spread but also any potential shift in the process mean. It is calculated as:\[ PCR_k = \frac{1}{3} \min\left(\frac{USL - \mu}{\sigma}, \frac{\mu - LSL}{\sigma}\right) \]PCR_k considers both the upper and lower bound of the process performance concerning the mean. This makes it suitable for understanding how off-centered the process might be. For exercise (a), when \( \mu = 100 \), the PCR_k calculates to:\[ PCR_k = 1.11 \]This indicates a centered and capable process. However, in part (b) with mean adjusted to 106, PCR_k evaluates as:\[ PCR_k = 0.78 \]This lower value signifies that though the process capability, as highlighted by PCR, remains unaltered (1.11), the mean shift decreases its effectiveness due to higher defect probabilities, particularly at the upper specification limit.

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Most popular questions from this chapter

An article in Quality Engineering ["Is the Process Capable? Tables and Graphs in Assessing \(C_{\text {rm }} "(1992,\) Vol. 4(4)\(]\). Considered manufacturing data. Specifications for the outer diameter of the hubs were \(60.3265 \pm 0.001 \mathrm{~mm}\). A random sample with size \(n=20\) was taken and the data are shown in the following table: $$\begin{array}{cccc}\hline \text { Sample } & x & \text { Sample } & x \\\\\hline 1 & 60.3262 & 11 & 60.3262 \\\2 & 60.3262 & 12 & 60.3262 \\\3 & 60.3262 & 13 & 60.3269 \\\4 & 60.3266 & 14 & 60.3261 \\\5 & 60.3263 & 15 & 60.3265 \\\6 & 60.3260 & 16 & 60.3266 \\\7 & 60.3262 & 17 & 60.3265 \\\8 & 60.3267 & 18 & 60.3268 \\\9 & 60.3263 & 19 & 60.3262 \\\10 & 60.3269 & 20 & 60.3266\end{array}$$ (a) Construct a control chart for individual measurements. Revise the control limits if necessary. (b) Compare your chart in part (a) to one that uses only the last (least significant) digit of each diameter as the measurement. Explain your conclusion. (c) Estimate \(\mu\) and \(\sigma\) from the moving range of the revised chart and use this value to estimate \(P C R\) and \(P C R_{k}\) and interpret these ratios.

Cover cases for a personal computer are manufactured by injection molding. Samples of five cases are taken from the process periodically, and the number of defects is noted. Twenty-five samples follow: $$\begin{array}{cccc}\hline \text { Sample } & \text { No. of Defects } & \text { Sample } & \text { No. of Defects } \\\\\hline 1 & 3 & 14 & 8 \\\2 & 2 & 15 & 0 \\\3 & 0 & 16 & 2 \\\4 & 1 & 17 & 4 \\\5 & 4 & 18 & 3 \\\6 & 3 & 19 & 5 \\\7 & 2 & 20 & 0 \\\8 & 4 & 21 & 2 \\\9 & 1 & 22 & 1 \\\10 & 0 & 23 & 9 \\\11 & 2 & 24 & 3 \\\12 & 3 & 25 & 2 \\\13 & 2 & &\end{array}$$ (a) Using all the data, find trial control limits for a \(U\) chart for the process. (b) Use the trial control limits from part (a) to identify out-ofcontrol points. If necessary, revise your control limits. (c) Suppose that instead of samples of five cases, the sample size was 10. Repeat parts (a) and (b). Explain how this change alters your responses to parts (a) and (b).

An automatic senson measures the diameter of holes in consecutive order. The results of measuring 25 holes follow. $$\begin{array}{cccc}\text { Sample } & \text { Diameter } & \text { Sample } & \text { Diameter } \\\\\hline 1 & 9.94 & 14 & 9.99 \\\2 & 9.93 & 15 & 10.12 \\\3 & 10.09 & 16 & 9.81 \\\4 & 9.98 & 17 & 9.73 \\\5 & 10.11 & 18 & 10.14 \\\6 & 9.99 & 19 & 9.96 \\\7 & 10.11 & 20 & 10.06 \\\8 & 9.84 & 21 & 10.11 \\\9 & 9.82 & 22 & 9.95 \\\10 & 10.38 & 23 & 9.92 \\\11 & 9.99 & 24 & 10.09 \\\12 & 10.41 & 25 & 9.85 \\\13 & 10.36 & &\end{array}$$ (a) Estimate the process standard deviation. (b) Set up a CUSUM control procedure, assuming that the target diameter is 10.0 millimeters. Does the process appear to be operating in a state of statistical control at the desired target level?

Suppose that an \(\bar{X}\) chart is used to control a normally distributed process and that samples of size \(n\) are taken every \(n\) hours and plotted on the chart, which has \(k\) -sigma limits. (a) Find a general expression for the expected number of samples and time that is taken until a false signal is generated. (b) Suppose that the process mean shifts to an out-of-control state, say \(\mu_{1}=\mu_{0}+\delta \sigma\). Find an expression for the expected number of samples that is taken until a false action is generated. (c) Evaluate the in-control ARL for \(k=3 .\) How does this change if \(k=2 ?\) What do you think about the use of 2-sigma limits in practice? (d) Evaluate the out-of-control ARL for a shift of 1 sigma, given that \(n=5\).

The \(P C R\) for a measurement is 1.5 and the control limits for an \(\bar{X}\) chart with \(n=4\) are 24.6 and 32.6 . (a) Estimate the process standard deviation \(\sigma\). (b) Assume that the specification limits are centered around the process mean. Calculate the specification limits.

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